Class 11 Commerce Maths - Complex Numbers And Quadratic Equations

If the roots of the equations $ax^{2}+2bx+c=0$ and $bx^{2}-2\sqrt{ac}x+b=0$ are simultaneously real, then prove that $b^{2}=ac$ .

Let

$D_{1}$ and

$D_{2}$ be the discriminants of the two equations, then

$D_{1}\geq 0$ and

$D_{2}\geq 0$

$\Rightarrow 4b^{2}-4ac\geq 0$ and

$4ac-4b^{2}\geq 0$

$\Rightarrow b^{2}\geq ac$ and

$ac\geq b^{2}$

$\Rightarrow b^{2}=ac$

Determine the nature of the roots of the equation $4x^2 - 4x + 1 = 0$

Consider the equation

$4x^2 - 4x + 1 = 0$
This is in the form of

$ax^2 + bx + c = 0$ . The co-efficient are

$a = 4, b = - 4, c = 1$

$\triangle = b^2 - 4ac, \triangle (-4)^2 - 4(4)(1), \triangle = 16 - 16$

$\therefore \triangle = 0$ . Therefore, roots are real and equal

Find the nature of the roots of the following:
${x}^{2}+x+1=0$

$x^{2}+x+1=0$

$ax^{2}+bx+c=0$ form,

$a=1,b=1,c=1$

Discriminant

$b^{2}-4ac=-3$ , which is

$< 0$

Hence, the root are complex.

If $b^2 - 4ac \geq 0$ , then write the nature of roots of a quadratic equation $ax^2 + bx + c = 0$ .

For

$ax^2+bx+c=0$ , roots of the equation are represented by,

$x=\dfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$b^2 - 4ac \geq 0$ , then the roots are real.

Find the no. of solutions of ${x}^{2}+2x+2=0$ .

$\Rightarrow$

$x^2+2x+2=0$

$\Rightarrow$

$a=1,\,b=2,\,c=2$

$\Rightarrow$

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$=\dfrac{-2\pm\sqrt{(2)^2-4(1)(2)}}{2(1)}$

$=\dfrac{-2\pm\sqrt{4-8}}{2}$

$=\dfrac{-2\pm\sqrt{-4}}{2}$

$=\dfrac{-2\pm 2i}{2}$

$=-1\pm i$

$\Rightarrow$

$x=-1+i$ and

$x=-1-i$

For what positive values of 'm', roots of the equation $x^2 + mx + 4 = 0$ are equal

Consider the equation

$x^2 + mx + 4 = 0$
This is in the form

$ax^2 + bx + c = 0$ . The co-efficients are

$a = 1, b = m, c = 4$

$\triangle = b^2 - 4ac, \triangle = m^2 - 4(1)(4), \triangle = m^2 - 16$
If roots are equal, then

$\triangle = 0$

$\therefore m^2 - 16 = 0, m^2 = 16, m = \pm \sqrt{16} \therefore m = 4$

What is the nature of the roots of the quadratic equation $a{ x }^{ 2 }+bx+c=0$ if
(a) ${ b }^{ 2 }-4ac=0$
(b) ${ b }^{ 2 }-4ac<0$

(a) If

${ b }^{ 2 }-4ac=0$ , then roots are equal.
(b) If

${ b }^{ 2 }-4ac<0$ , then roots are complex or imaginary.

In the given, determine whether the given quadratic equation has real roots and if so, find the roots.
$25{ x }^{ 2 }+20x+7=0$

The given equation

$25{ x }^{ 2 }+20x+7=0$

Here,

$a=25,b=20,c=7$

$\therefore D={ b }^{ 2 }-4ac={ 20 }^{ 2 }-4\times 25\times 7=-300<0\quad$

So, the given equation does not have real roots.

Determine the nature of the roots of the given quadratic equation
$2{ x }^{ 2 }+5x+5=0\quad$

The given equation is

$2{ x }^{ 2 }+5x+5=0\quad$

Here,

$a=2,b=5,c=5$

$\quad \therefore D={ b }^{ 2 }-4ac={ 5 }^{ 2 }-4\times 2\times 5=-15\quad$

Since,

$D< 0$ roots of the given equation are not real

Determine the nature of the roots of the given quadratic equation
${ x }^{ 2 }-4x+4=0$

The given equation is

${ x }^{ 2 }-4x+4=0$

$\text{Comparing given equation with standard quadratic equation } ax^2+bx+c =0$

We get

$a=1,b=-4 \text{ and }c=4$

$\textbf{Step 1: Find discriminant D of the given equation}$

$\text {We have } D={ b }^{ 2 }-4ac$

$\Rightarrow D ={ (-4) }^{ 2 }-4\times 1\times 4=0$

$\textbf{Step 2: Conclusion }$

Since

$D=0 \Rightarrow$ roots of the given equation are real and equal.

Determine the nature of the roots of the follwoing quadratic equation :
$2x^2-6x+3=0$

$2x^2-6x+3=0$

This is of the form

$ax^2+bx+c=0$ , where

$a=2,b=-6,c=3$ .

$D=b^2-4ac=36-24=12>0$

Therefore, the given equation has real unequal roots.

Determine the nature of roots of the given quadratic equation $3x^{2} + 2\sqrt {5}x - 5 = 0$ .

Given,

$3x^2+2\sqrt 5 -5=0$

$a=3,b=2\sqrt {5},c=-5$

Discriminant:

$=b^2-4ac$

$=(2\sqrt {5})^2-4(3)(-5)$

$=20+60=80>0$

Since,

$b^2-4ac>0\Rightarrow$ roots are real and distinct.

$x^{2}+1$ has no zeros. Why?

$x^2+1$ , has discriminan

$= b^2-4ac = 0^2-4(1)(1) = -4<0$

So, this has no real roots,

Hence it has no zeroes,

Find $a$ so that roots of $x ^ { 2 } + 2 ( 3 a + 5 ) x + 2 \left( 9 a ^ { 2 } + 25 \right) = 0$ are real.

For roots to be real,

$D\geq$ 0

So ,

$4(3a+5)^{2}-4\times 1\times 2(9a^{2}+25)\geq 0$

$\rightarrow 9a^{2}+25+30a-18a^{2}-50\geq 0$

$\rightarrow -9a^{2}+30a-25\geq0$

$\rightarrow 9a^{2}-30a+25\leq0$

$\rightarrow (3a-5)^{2}\leq0$

But square of a real number can't be less than 0

So only equality can hold

Thus a=

$\dfrac{5}{3}$

Find the values of $m$ for which the quadratic equation ${x}^{2}-m(2x-8)-15=0$ has (i) equal roots; (ii) both roots positive

(i) For equal roots,

$b^2-4ac=0$ ,

$4m^2-4(8m-15)=0$

$4m^2-8m+15=0$

$m^2-8m+15=0$

$(m-3)(m-5)=0$

$m=3,m=5$

(ii) For roots positive

$b^2-4ac>0$

$m\in(3,\infty), m\in(5,\infty)$

Find the discriminant of the following quadratic equations and discuss the nature of the roots .
$6 x ^ { 2 } - 13 x + 6 = 0$
2. $\sqrt { 6 } x ^ { 2 } - 5 x + \sqrt { 6 } = 0$
3. $24 x ^ { 2 } - 17 x + 3 = 0$
4. $x ^ { 2 } + 2 x + 4 = 0$
5. $x ^ { 2 } + x + 1 = 0$
6. $x ^ { 2 } - 3 \sqrt { 3 } x - 30 = 0$

$1) 6x^2-13x+6=0$

$D=b^2-4ac = (-13)^2-4 \times 6 \times 6=169-144=25>0$

Hence, the roots are real and unequal.

$2) \sqrt{6}x^2-5x+\sqrt{6}=0$

$D=b^2-4ac=(-5)^2-4\times \sqrt{6} \times \sqrt{6}=25-24=1>0$

Hence, the roots are real and unequal.

$3) 24x^2-17x+3=0$

$D=b^2-4ac=(-17)^2-4\times 24 \times 3 = 289-288=1>0$

Hence, the roots are real and unequal.

$4) x^2+2x+4=0$

$D=b^2-4ac=4-4\times 1 \times 4=4-16=-12<0$

Hence, it has no real roots.

$5) x^2+x+1=0$

$D=b^2-4ac=1-4\times 1 \times 1=1-4=-3<0$

Hence, it has no real roots.

$6) x^2-3\sqrt{3}x-30=0$

$D=b^2-4ac=(-3\sqrt{3})^2-4\times 1 \times (-30)=27+120=147>0$

Hence, the roots are real and unequal.

Without solving, comment upon the nature of roots of each of the following equations:
$6 x^{2}-13 x+4=0$

Given quadratic equation,

$6 x^{2}-13 x+4=0$ Here,

$a=6, b=-13$ and

$c=4$
So, the Discriminant (D)

$=b^{2}-4 a c$

$D=(-13)^{2}-4(6)(4)$

$=169-48$

$=121$
As

$\mathrm{D}>0,$ the roots of the equation is real and unequal.

Without solving, comment upon the nature of roots of each of the following equations:
$2 x^{2}+8 x+9=0$

Given quadratic equation,

$2 x^{2}+8 x+9=0$
Here,

$a=2, b=8$ and

$c=9$
So, the Discriminant (D)

$=b^{2}-4 a c$

$D=(8)^{2}-4(2)(9)$

$=64-72$

$=-8$
As

$D<0,$ the equation has no roots.

Without solving, comment upon the nature of roots of each of the following equations:
$25 x^{2}-10 x+1=0$

Given quadratic equation,

$25 x^{2}-10 x+1=0$
Here,

$a=25, b=-10$ and

$c=1$
So, the Discriminant (D)

$=b^{2}-4 a c$

$D=(-10)^{2}-4(25)(1)$

$=100-100$

$=0$
As

$D=0,$ the roots of the equation are real and equal.

Without solving, comment upon the nature of roots of each of the following equations:
$7 x^{2}-9 x+2=0$

Given quadratic equation,

$7 x^{2}-9 x+2=0$ Here,

$a=7, b=-9$ and

$c=2$
So, the Discriminant (D)

$=b^{2}-4 a c$

$D=(-9)^{2}-4(7)(2)$

$=81-56$

$=25$
As

$\mathrm{D}>0,$ the roots of the equation is real and unequal.

If the equation $(1+m^{2})x^{2}+2mcx+(c^{2}-a^{2})=0$ has equal roots, then prove that $c^{2}=a^{2}(1+m^{2})$ .

Given Equation

$=(1+m^2)x^2+2mc x+(c^2-a^2)=0$

Given condition = Roots are equal

Hence

$D=0$ [For Roots to be equal]

$D=(2mc)^2-4(1+m^2)(c^2-a^2)$

$\Rightarrow 4m^2c^2-4(c^2+m^2c^2-a^2-m^2a^2)=0$

$\Rightarrow 4m^2c^2-4c^2-4m^2c^2+4a^2+4m^2a^2=0$

$\Rightarrow -4c^2+4a^2+4m^2a^2=0$

$\Rightarrow -c^2+a^2+m^2a^2=0$

$\Rightarrow \boxed{c^2=a^2+m^2a^2}$

Required condition

$\Rightarrow c^2=a^2+a^2m^2$

$c^2=a^2(1+m^2)$

Show that the equation $2(a^{2}+b^{2})x^{2}+2(a+b)x+1=0$ has no real roots, when $a\neq b$

$D=b^{2}-4ac=0$

$=2\left [ 2(a+b) \right ]^{2}-2\left [ 2(a^{2}+b^{2}) \right ]\left [ 1 \right ]$

$=4(a+b)^{2}-8(a^{2}+b^{2})$

$=4a^{2}+4b^{2}+8ab-8a^{2}-8b^{2}$

$=-4a^{2}-4b^{2}+8ab=-4(a-b)^{2}$
must be less than zero
Hence no real roots.

Prove that both the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are real but they are equal only when $a=b=c$ .

$3x^{2}-2x(a+b+c)+(ab+bc+ca)=0$

Let

$D$ be the discriminant

$D=4(a+b+c)^{2}-12(ab+bc+ca)$

$=4\left [ (a+b+c)^{2}-3(ab+bc+ca) \right ]$

$=2\left [ 2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca \right ]$

$=2\left [ (a-b)^{2}+(b-c)^{2}+(c-a)^{2} \right ]$

$D=0$ , then

$(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$

$\Rightarrow a-b=0 ; b=c; c=a$

$a=b=c$

$y^2 - 7y + 2 = 0$

We know that while finding the root of a quadratic equation

$ax^2+bx+c=0$ by quadratic formula

$x=\frac { -b±\sqrt { b^{ 2 }-4ac } }{ 2a }$ ,

$b^2-4ac>0$ , then the roots are real and distinct

$b^2-4ac=0$ , then the roots are real and equal and

$b^2-4ac<0$ , then the roots are imaginary.

Here, the given quadratic equation

$y^2-7y+2=0$ is in the form

$ax^2+bx+c=0$ where

$a=1,b=-7$ and

$c=2$ , therefore,

$b^{ 2 }-4ac=(-7)^{ 2 }-(4\times 1\times 2)=49-8=41>0$

Since

$b^2-4ac>0$

Hence the roots are real and distinct.

Determine the nature of the roots of the equation $2x^2 - 5x - 1 = 0$

This is in the form of

$ax^2 + bx + c = 0$ . The co-efficient are

$a = 2, b = -5, c = -1$

$\triangle = b^2 - 4ac, \triangle = (-5)^2 - 4(2) (-1), \triangle = 25 + 8, \triangle = 33$

$\therefore \triangle > 0$ . Therefore, roots are real and distinct

Prove that the roots of the equation $(a-b+c)x^2+2(a-b)x+(a-b-c)=0$ are rational numbers for all real numbers a and b and for all rational c.

Let the given equation be of the form

$Ax^2+Bx+C=0$ .

Then,

$A= a-b+c, B= 2(a-b)$ and

$C = a- b - c$ .

Now, the discriminant of

$Ax^2+ Bx+ C=0$ is

$B^2-4AC=[2(a-b)]^2-4(a-b+c)(a-b-c)$

$=4(a-b)^2-4[(a-b)+c][(a-b)-c]$

$= 4(a-b)^2-4[(a-b)^2-c^2]$

$\Delta = 4(a-b)^2-4(a-b)^2+4c^2=4c^2$ , a perfect square.

Therefore,

$\Delta > 0$ and it is a perfect square.

Hence, the roots of the given equation are rational numbers.

Determine the nature of the roots of the given quadratic equation
${ x }^{ 2 }+x+1=0$

$\text{The given equation is }$

${ x }^{ 2 }+x+1=0$

$\textbf{ Step 1 : On comparing the given equation with }$

$\boldsymbol{ax^2 +bx +c =0 }$

$\text { Here, a=1,b=1,c=1 }$

$\textbf{Step 2 : Evaluating the discriminant of the given quadratic equation.}$

$\quad \therefore D={ b }^{ 2 }-4ac$

$={ 1 }^{ 2 }-4\times 1\times 1=-3$

$\text{Therefore, the roots of the given equation are not real.}$

In the given, determine whether the given quadratic equation has real roots and if so, find the roots.
$9{ x }^{ 2 }+7x-2=0$

The given equation is

$9{ x }^{ 2 }+7x-2=0$

Here

$a=9,b=7,c=2$

$\therefore D={ b }^{ 2 }-4ac={ (7) }^{ 2 }-4\times 9\times -2=121>0\quad$

So, the given equation has real roots given by

$\alpha =\cfrac { -b+\sqrt { D } }{ 2a } =\cfrac { -7+\sqrt { 121 } }{ 2\times 9 } =\cfrac { 2 }{ 9 }$

$\beta =\cfrac { -b-\sqrt { D } }{ 2a } =\cfrac { -7-\sqrt { 121 } }{ 2\times 9 } =-1$

Number of real roots of ${x} + \sqrt{1- x^2}={5}$

Solution:-

$x + \sqrt{1 - {x}^{2}} = 5 \; \left\{ Given \right\}$

$\Rightarrow \; x - 5 = -\sqrt{1 - {x}^{2}}$

Squaring both sides, we have

${\left( x - 5 \right)}^{2} = {\left( -\sqrt{1 - {x}^{2}} \right)}^{2}$

${x}^{2} + 25 - 2 \times 5 \times x = 1 - {x}^{2}$

$\Rightarrow \; {x}^{2} + 25 - 10x = 1 - {x}^{2}$

$\Rightarrow \; 2{x}^{2} - 10x + 24 = 0$

$\therefore \; D = {b}^{2} - 4ac = {\left( -10 \right)}^{2} - 4 \times 2 \times 24 = -92$

Since

$D < 0$ , there will be no real roots.

State whether the following quadratic equation have too distinct equal roots:
a) $( x - \sqrt { 2 } ) ^ { 2 } - 2 ( x + 1 ) = 0 \quad { b) } \sqrt { 2 } x ^ { 2 } - \frac { 3 } { \sqrt { 2 } } x + \frac { 1 } { \sqrt { 2 } } = 0$

(a)

$(x-\sqrt{2})^{2}-2(x+1) = 0.$

$x^{2}-2\sqrt{2x}+2-2x-2 = 0$

$x^{2}-2x(\sqrt{2}+1) = 0\Rightarrow . x = 0, x = 2\sqrt{2}+2$

$x(x-2(\sqrt{2+1})) = 0$

it has two real and distinct roots.

(b)

$\displaystyle \sqrt{2x^{2}}-\frac{3}{\sqrt{2}}x+\frac{1}{\sqrt{2}} = 0$

$x^{2}-\dfrac{3}{2}x+\dfrac{1}{2} = 0$

$\Rightarrow 2x^{2}-3x+1 = 0$

$2x^{2}-2x-x+1 = 0$

$2x(n-1)-1(n-1) = 0$

$x = \dfrac{1}{2},1$

it has two real and distinct real roots.

1100766_1173383_ans_efa9e65c643e47209f1327787a03d4eb.jpg

Find the nature of the roots of the following quadratic equations . If the real roots exit. find them:
1) $2{x^2} - 3x + 5 = 0$
2) $3{x^2} - \root 4 \of 3 x + 4 = 0$

Solution:-

Given quadratic equations

(i)

$2x^{2}-3x+5 = 0$

comparing with the general equation of quadratic

$ax^{2}+bx+c,$ we get

$-a = 2$

$b = -3$

$c = 5$

$\therefore D = b^{2}-4ac = 9-40 < 0$

$\Rightarrow$ Imaginary roots

(ii)

$3x^{2}-\sqrt[4]{3}x+4 = 0$

comparing with general quadratic equation

$a = 3$

$b = \sqrt[4]{3}$

$c = 4$

$\therefore D = b^{2}-4ac = \sqrt{3}-48< 0$

$\Rightarrow$ Imaginary roots

Determine the set of values of $k$ for which the given quadratic equation has real roots:
$2x^{2}+kx-4=0$

$2x^2+kx-4=0$

Here discriminant is

$D=k^2-4(-4)(2)$

For real root

$D\ge0$

Hence

$k^2+32\ge0$

$\implies k^2\ge-32$

Hence set of values are

$\{k:k^2\ge-32\}$

Show that the equation $x^{2}+ax+1=0$ has real and distinct roots for all real values of $a$ .

$D=b^2-4ac$

$D=a^2-4(1)(1)=a^2-4$

for all the real values of

$a>2\, and\, a<-2$

$D>0$

Hence

$x^2+ax+1=0$ has real and distinct roots for all the real values of a

Write the nature of the roots of the folliowing quadratic equations
$\quad { 2x }^{ 2 }+6x+3=0$

1292610_1282215_ans_90f396f69ec74907b264b870b6e29ba0.jpg

How many roots of the following quadratic equations are real:
$4x^{2}-4a^{2}x+(a^{4}-b^{4})=0$

On comparing the given quadratic equation with

$ax^2+bx+c=0$ we get,

$a=4,\ b=-4a^2$ and

$c=a^4-b^4$

So,

$D=b^2-4ac$

$= {\left( { - 4{a^2}} \right)^2} - 4\left( 4 \right)\left( {{a^4} - {b^4}} \right)$

$= 16{a^4} - 16\left( {{a^4} - {b^4}} \right)$

$= 16{a^4} - 16{a^4} + 16{b^4}$

$= 16{b^4}$

$D>0,$ this implies that the given quadratic equation has two real and distinct roots.

$(k+1)x^{2}-2(3k+1)x+8k+1=0$ has equal roots

Condition to have equal roots

$b^2=4ac\\4(3k+1)^2=4(k+1)(8k+1)\\9k^2+6k+1=8k^2+9k+1\\k^2=3k\\k=0,3$

Find the value of k for which the Q.E $kx^2+1-2(k-1)x+x^2=0$ has equal roots.

Given :

$kx^{2}+1-2(k-1)x+x^{2}=0$

$\Rightarrow kx^{2}+x^{2}-2(k-1)+1 =0$

$\Rightarrow x^{2}(k+1)-2(k-1)+1=0$

Here It is in the form ob

$ax^{2}+bx+c=0$

$a=(k+1) b=-2(k-1) c=1$

for real and equal roots;

$D = 0$

$\Rightarrow b^{2}-4ac=0$

$\Rightarrow [-2(k-1)]^{2}-4(k+1).1=0$

$\Rightarrow4(k^{2}+1-2k)-4k-4=0$

$\Rightarrow 4k^{2}+4-8k-4k-4=0$

$\Rightarrow 4k^{2}-12k=0$

$\Rightarrow 4k(k-3)=0$

$4 k = 0$ or

$k-3 = 0$

$k = 0/4$

$k = 3$

$k = 0$

$\therefore$

$k= 0,3.$

Cheek whether the quadratic equation $\left( { a }^{ 2 }+{ b }^{ 2 } \right) { x }^{ 2 }+2\left( ac+bd \right) x+{ c }^{ 2 }+{ d }^{ 2 }=0$ has real roots.

$(a^2+b^2)x^2+2(ac+bd)x+(c^2+d^2)=0$

$D=b^2-4ac$

$D=4(ac+bd)^2-4(a^2+b^2)(c^2+d^2)$

$D=4(a^2c^2+b^2d^2+2abcd)-4(a^2c^2+a^2d^2+b^2c^2+b^2d^2)$

$D=4(2abcd-a^2d^2-b^2c^2)$

$D=-4(ad-bc)^2$

Since

$(ad-bc)^2$ is always positive.

$\therefore$ Roots are not real.

They are imaginary.

1208526_1450511_ans_297a7cb21bcb47adafc10e1cfce1692b.jpg

Find the nature of the roots of the following quadratic equation:
1. $3{x^2} + 7x + 2 = 0$
2. ${\left( {x + 3} \right)^2} + 4x + 7 = 0$
3. $2{x^2} - 8x + 34 = 0$
4. ${x^2} + 2\sqrt {3x} - 1 = 0$

$3{ x }^{ 2 }+7x+2=0$

$\triangle ={ b }^{ 2 }-4ac={ 7 }^{ 2 }-4\left( 2 \right) \left( 3 \right) =49-24=25>0$

$\therefore$ Roots are real and positive

${ \left( x+3 \right) }^{ 2 }+4x+7=0$

${ x }^{ 2 }+9+6x+4x+7=0$

${ x }^{ 2 }+10x+16=0$

$\triangle ={ b }^{ 2 }-4ac=100-4\left( 16 \right) =36={ 6 }^{ 2 }>0$

$\therefore$ Roots are real and positive

$2{ x }^{ 2 }-8x+34=0$

$\triangle ={ 8 }^{ 2 }-4\left( 2 \right) \left( 34 \right) <0$

$\therefore$ Roots are imaginary

${ x }^{ 2 }+2\sqrt { 3 } x-1=0$

$\triangle ={ \left( 2\sqrt { 3 } \right) }^{ 2 }+4\left( 1 \right) =12+4=16>0$

Roots are real and positive.

Without solving, comment upon the nature of roots of each of the following equations:
$x^{2}-a x-b^{2}=0$

Given quadratic equation,

$x^{2}-a x-b^{2}=0$
Here,

$a=1, b=-a$ and

$c=-b^{2}$
So, the Discriminant

$(\mathrm{D})=\mathrm{b}^{2}-4 \mathrm{ac}$

$D=(a)^{2}-4(1)\left(-b^{2}\right)$

$=a^{2}+4 b^{2}$

$a^{2}+4 b^{2}$ is always positive value.
Thus

$D>0$ , and the roots of the equation is real and unequal

Without solving, comment upon the nature of roots of each of the following equations:
$x^{2}+2 \sqrt{3} x-9=0$

Given quadratic equation,

$x^{2}+2 \sqrt{3} x-9=0$
Here,

$a=1, b=2 \sqrt{3}$ and

$c=-9$
So, the Discriminant

$(\mathrm{D})=\mathrm{b}^{2}-4 \mathrm{ac}$

$D=(2 \sqrt{3})^{2}-4(1)(-9)$

$=12+36$

$=48$
As

$\mathrm{D}>0,$ the roots of the equation are real and unequal.

Find the nature of roots of following quadratic equations.
$2x^2+x-1=0$

Given equation

$2x^2+x-1=0$
Comparing with

$ax^2+bx+c=0$

$a=2, b=1$ and

$c=-1$
Discriminant

$(D)=b^2-4ac$

$=(1)^2-4 \times 2 \times (-1)$

$=1+8$

$=9 > 0$
Thus roots of equation are real and distinct

Write nature of roots by general form of discriminant of equation $ax^2+bx+c=0$ .

Nature of roots of quadratic equation

$ax^2+bx+c=0, a=0$ depends on its discriminant value,

$D=b^2-4ac$
(i) If

$D=b^2-4ac > 0$ , then roots will be real and distinct.
If

$\alpha$ and

$\beta$ are roots of the equation then

$\alpha =\dfrac {-b+\sqrt{b^2-4ac}}{2a}$ and

$\beta =\dfrac {-b-\sqrt{b^2-4ac}}{2a}$
(ii) If

$D=b^2-4ac=0$ , then roots will be real and same i.e.

$\alpha =-\dfrac {b}{2a}=\beta$
(iii)

$D=b^2-4ac < 0$ , then roots will not be real i.e. root will be imaginary.

Find the value of $k$ , for which following quadratic equations have real and equal roots
$2x^2+kx+3=0$

Given equation

$2x^2+kx+3=0$
Comparing with

$ax^2+bx+c=0$

$a=2, b=k$ and

$c=3$
Discriminant

$(D)=b^2-4ac$

$=k^2-4 \times 2 \times 3$

$=k^2-24$
For equal roots

$D=0$

$\Rightarrow k^2-24=0$

$\Rightarrow k^2=24$

$\Rightarrow k=\pm \sqrt{24}=\pm 2 \sqrt 6$
Thus

$k=\pm 2 \sqrt 6$

Find the value of $k$ , for which following quadratic equations have real and equal roots
$x^2-2(k+1)x+k^2=0$

Given equation

$x^2-2(k+1)x+k^2=0$
Comparing with

$ax^2+bx+c=0$

$a=1, b=-2(k+1)$ and

$c=k^2$
Discriminant

$(D)=b^2-4ac$

$=\left\{-2(k+1)\right\}^2-4 \times 1 \times k^2$

$=4(k^2+2k+1)-4k^2$

$4k^2+8k+4-4k^2$

$=8k+4$
For equal roots

$D=0$

$\Rightarrow 8k+4=0$

$\Rightarrow 8k=-4$

$\Rightarrow k=\dfrac{-1}{2}$
Thus

$k=\dfrac{-1}{2}$

For quadratic equation $ax^2+bx+c=0, a\neq 0$ , at which nature of roots depends?

Discriminant

$(D)=b^2 -4ac$

Describe nature of roots of quadratic equation $ax^2 +bx+c=0, a\neq 0$

(i) If

$(b^2 -4ac) > 0$ \, then roots will be real and distinct.
(ii) If

$(b^2 -4ac) = 0$ \, then roots will be equal and real.
(iii) If

$(b^2 -4ac) < 0$ \, then roots will be imaginary.

Without solving the following quadratic equation, find the greater value of $p$ for which the given equation has real and equal roots. $x^2+(p-3)x+p=0$ .

Given equation

$x^2+(p-x)x+p=0$

$\because$ Roots are real and equal, then

$b^2-4ac=0$

Here we compare the coefficients of a, b and c with the equation

$ax^2+bx+c=0$ .

$a=1, b=p-3$ and

$c=p$

Now putting the values of a, b and c in equation

$(p-3)^2-4\times 1\times p=0$

$\Rightarrow p^2+9-6p-4p=0$

$\Rightarrow p^2+9-10p=0$

$\Rightarrow p^2-10p+9=0$

$\Rightarrow p^2-9p-p+9=0$

$\Rightarrow p(p-9)-1(p-9)=0$

$\Rightarrow (p-9)(p-1)=0$

Hence,

$p=9$ or

$1$ .

Prove that the roots of the equation $bx^{2} + (b - c) x + b(b - c - a) = 0$ are real if those of $ax^{2} + 2bx + b = 0$ are imaginary and vice versa.

For roots to be real

${ \left( b-c \right) }^{ 2 }-4.b.\left( b-c-a \right) >0$

$\left( { b }^{ 2 }+{ c }^{ 2 }-2bc \right) -4\left( { b }^{ 2 }-bc-ab \right) >0$

Solving above equation,

$-3{ b }^{ 2 }+2bc+{ c }^{ 2 }+4ab>0$

$\left( 2bc+{ c }^{ 2 } \right) -3{ b }^{ 2 }+4ab>0$

Making square of

$\left( b-c \right)$

${ \left( b-c \right) }^{ 2 }-4b\left( b-c-a \right) >0$

Solving above

${ \left( b-c \right) }^{ 2 }-4{ b }^{ 2 }+4bc+4ab>0\\ { \left( b+c \right) }^{ 2 }-4{ b }^{ 2 }+4ab>0\\ { \left( b+c \right) }^{ 2 }>4\left( { b }^{ 2 }-4ab \right)$

Now as per given condition, for roots to be imaginary

$4{ b }^{ 2 }-4ab<0\\ 4\left( { b }^{ 2 }-ab \right) <0\\ 4b\left( b-a \right) <0$

$\left( a-b \right) >0$

If $a$ and $c$ are odd prime numbers and $a x^{2}+b x+c=0$ has rational roots where $b \in I .$ Prove that one root of the equation will be independent of $a, b, c$

1947826_1746125_ans_7502dc64fc6f4dccbd88b138e62d8f5a.jpg

Complex Numbers And Quadratic Equations - Class 11 Commerce Maths - Extra Questions

If the roots of the equations ax2+2bx+c=0ax^{2}+2bx+c=0 and bx2−2√acx+b=0bx^{2}-2\sqrt{ac}x+b=0 are simultaneously real, then prove that b2=acb^{2}=ac.

Determine the nature of the roots of the equation 4x2−4x+1=04x^2 - 4x + 1 = 0

Find the nature of the roots of the following:x2+x+1=0{x}^{2}+x+1=0

If b2−4ac≥0b^2 - 4ac \geq 0, then write the nature of roots of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0.

Find the no. of solutions of x2+2x+2=0{x}^{2}+2x+2=0.

For what positive values of 'm', roots of the equation x2+mx+4=0x^2 + mx + 4 = 0 are equal

What is the nature of the roots of the quadratic equation ax2+bx+c=0a{ x }^{ 2 }+bx+c=0 if(a) b2−4ac=0{ b }^{ 2 }-4ac=0(b) b2−4ac<0{ b }^{ 2 }-4ac<0

In the given, determine whether the given quadratic equation has real roots and if so, find the roots.25x2+20x+7=025{ x }^{ 2 }+20x+7=0

Determine the nature of the roots of the given quadratic equation2x2+5x+5=02{ x }^{ 2 }+5x+5=0\quad

Determine the nature of the roots of the given quadratic equationx2−4x+4=0{ x }^{ 2 }-4x+4=0

Determine the nature of the roots of the follwoing quadratic equation :2x2−6x+3=02x^2-6x+3=0

Determine the nature of roots of the given quadratic equation 3x2+2√5x−5=03x^{2} + 2\sqrt {5}x - 5 = 0.

x2+1x^{2}+1 has no zeros. Why?

Find aa so that roots of x2+2(3a+5)x+2(9a2+25)=0x ^ { 2 } + 2 ( 3 a + 5 ) x + 2 \left( 9 a ^ { 2 } + 25 \right) = 0 are real.

Find the values of mm for which the quadratic equation x2−m(2x−8)−15=0{x}^{2}-m(2x-8)-15=0 has (i) equal roots; (ii) both roots positive

Without solving, comment upon the nature of roots of each of the following equations:6x2−13x+4=0 6 x^{2}-13 x+4=0

Without solving, comment upon the nature of roots of each of the following equations:2x2+8x+9=0 2 x^{2}+8 x+9=0

Without solving, comment upon the nature of roots of each of the following equations:25x2−10x+1=0 25 x^{2}-10 x+1=0

Without solving, comment upon the nature of roots of each of the following equations:7x2−9x+2=0 7 x^{2}-9 x+2=0

If the equation (1+m2)x2+2mcx+(c2−a2)=0(1+m^{2})x^{2}+2mcx+(c^{2}-a^{2})=0 has equal roots, then prove that c2=a2(1+m2)c^{2}=a^{2}(1+m^{2}).

Show that the equation 2(a2+b2)x2+2(a+b)x+1=02(a^{2}+b^{2})x^{2}+2(a+b)x+1=0 has no real roots, when a≠ba\neq b

Prove that both the roots of the equation (x−a)(x−b)+(x−b)(x−c)+(x−c)(x−a)=0(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are real but they are equal only when a=b=ca=b=c.

Discuss the nature of roots of the given equation y2−7y+2=0y^2 - 7y + 2 = 0

Determine the nature of the roots of the equation 2x2−5x−1=02x^2 - 5x - 1 = 0

Prove that the roots of the equation (a−b+c)x2+2(a−b)x+(a−b−c)=0(a-b+c)x^2+2(a-b)x+(a-b-c)=0 are rational numbers for all real numbers a and b and for all rational c.

Determine the nature of the roots of the given quadratic equationx2+x+1=0{ x }^{ 2 }+x+1=0

In the given, determine whether the given quadratic equation has real roots and if so, find the roots.9{ x }^{ 2 }+7x-2=09{ x }^{ 2 }+7x-2=0

Number of real roots of {x} + \sqrt{1- x^2}={5}{x} + \sqrt{1- x^2}={5}

Find the nature of the roots of the following quadratic equations . If the real roots exit. find them:1) 2{x^2} - 3x + 5 = 02{x^2} - 3x + 5 = 02) 3{x^2} - \root 4 \of 3 x + 4 = 03{x^2} - \root 4 \of 3 x + 4 = 0

Determine the set of values of kk for which the given quadratic equation has real roots:2x^{2}+kx-4=02x^{2}+kx-4=0

Show that the equation x^{2}+ax+1=0x^{2}+ax+1=0 has real and distinct roots for all real values of aa.

Write the nature of the roots of the folliowing quadratic equations \quad { 2x }^{ 2 }+6x+3=0\quad { 2x }^{ 2 }+6x+3=0

How many roots of the following quadratic equations are real: 4x^{2}-4a^{2}x+(a^{4}-b^{4})=04x^{2}-4a^{2}x+(a^{4}-b^{4})=0

For what value/values of kk: (k+1)x^{2}-2(3k+1)x+8k+1=0(k+1)x^{2}-2(3k+1)x+8k+1=0 has equal roots

Find the value of k for which the Q.E kx^2+1-2(k-1)x+x^2=0kx^2+1-2(k-1)x+x^2=0 has equal roots.

Cheek whether the quadratic equation \left( { a }^{ 2 }+{ b }^{ 2 } \right) { x }^{ 2 }+2\left( ac+bd \right) x+{ c }^{ 2 }+{ d }^{ 2 }=0\left( { a }^{ 2 }+{ b }^{ 2 } \right) { x }^{ 2 }+2\left( ac+bd \right) x+{ c }^{ 2 }+{ d }^{ 2 }=0 has real roots.

Find the nature of the roots of the following quadratic equation:1.3{x^2} + 7x + 2 = 03{x^2} + 7x + 2 = 02.{\left( {x + 3} \right)^2} + 4x + 7 = 0{\left( {x + 3} \right)^2} + 4x + 7 = 03.2{x^2} - 8x + 34 = 02{x^2} - 8x + 34 = 04.{x^2} + 2\sqrt {3x} - 1 = 0{x^2} + 2\sqrt {3x} - 1 = 0

Without solving, comment upon the nature of roots of each of the following equations: x^{2}-a x-b^{2}=0 x^{2}-a x-b^{2}=0

Without solving, comment upon the nature of roots of each of the following equations: x^{2}+2 \sqrt{3} x-9=0 x^{2}+2 \sqrt{3} x-9=0

Find the nature of roots of following quadratic equations.2x^2+x-1=02x^2+x-1=0

Write nature of roots by general form of discriminant of equation ax^2+bx+c=0ax^2+bx+c=0.

Find the value of kk, for which following quadratic equations have real and equal roots2x^2+kx+3=02x^2+kx+3=0

Find the value of kk, for which following quadratic equations have real and equal rootsx^2-2(k+1)x+k^2=0x^2-2(k+1)x+k^2=0

For quadratic equation ax^2+bx+c=0, a\neq 0ax^2+bx+c=0, a\neq 0, at which nature of roots depends?

Describe nature of roots of quadratic equation ax^2 +bx+c=0, a\neq 0ax^2 +bx+c=0, a\neq 0

Without solving the following quadratic equation, find the greater value of pp for which the given equation has real and equal roots. x^2+(p-3)x+p=0x^2+(p-3)x+p=0.

Prove that the roots of the equation bx^{2} + (b - c) x + b(b - c - a) = 0bx^{2} + (b - c) x + b(b - c - a) = 0 are real if those of ax^{2} + 2bx + b = 0ax^{2} + 2bx + b = 0 are imaginary and vice versa.

If a a and c c are odd prime numbers and a x^{2}+b x+c=0 a x^{2}+b x+c=0 has rational roots where b \in I . b \in I . Prove that one root of the equation will be independent of a, b, c a, b, c

Class 11 Commerce Maths Extra Questions

Support mcqexams.com by disabling your adblocker.

If the roots of the equations $ax^{2}+2bx+c=0$ and $bx^{2}-2\sqrt{ac}x+b=0$ are simultaneously real, then prove that $b^{2}=ac$ .

Determine the nature of the roots of the equation $4x^2 - 4x + 1 = 0$

Find the nature of the roots of the following:
${x}^{2}+x+1=0$

If $b^2 - 4ac \geq 0$ , then write the nature of roots of a quadratic equation $ax^2 + bx + c = 0$ .

Find the no. of solutions of ${x}^{2}+2x+2=0$ .

For what positive values of 'm', roots of the equation $x^2 + mx + 4 = 0$ are equal

What is the nature of the roots of the quadratic equation $a{ x }^{ 2 }+bx+c=0$ if
(a) ${ b }^{ 2 }-4ac=0$
(b) ${ b }^{ 2 }-4ac<0$

In the given, determine whether the given quadratic equation has real roots and if so, find the roots.
$25{ x }^{ 2 }+20x+7=0$

Determine the nature of the roots of the given quadratic equation
$2{ x }^{ 2 }+5x+5=0\quad$

Determine the nature of the roots of the given quadratic equation
${ x }^{ 2 }-4x+4=0$

Determine the nature of the roots of the follwoing quadratic equation :
$2x^2-6x+3=0$

Determine the nature of roots of the given quadratic equation $3x^{2} + 2\sqrt {5}x - 5 = 0$ .

$x^{2}+1$ has no zeros. Why?

Find $a$ so that roots of $x ^ { 2 } + 2 ( 3 a + 5 ) x + 2 \left( 9 a ^ { 2 } + 25 \right) = 0$ are real.

Find the values of $m$ for which the quadratic equation ${x}^{2}-m(2x-8)-15=0$ has (i) equal roots; (ii) both roots positive

Without solving, comment upon the nature of roots of each of the following equations:
$6 x^{2}-13 x+4=0$

Without solving, comment upon the nature of roots of each of the following equations:
$2 x^{2}+8 x+9=0$

Without solving, comment upon the nature of roots of each of the following equations:
$25 x^{2}-10 x+1=0$

Without solving, comment upon the nature of roots of each of the following equations:
$7 x^{2}-9 x+2=0$

If the equation $(1+m^{2})x^{2}+2mcx+(c^{2}-a^{2})=0$ has equal roots, then prove that $c^{2}=a^{2}(1+m^{2})$ .

Show that the equation $2(a^{2}+b^{2})x^{2}+2(a+b)x+1=0$ has no real roots, when $a\neq b$

Prove that both the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are real but they are equal only when $a=b=c$ .

$y^2 - 7y + 2 = 0$

Determine the nature of the roots of the equation $2x^2 - 5x - 1 = 0$

Prove that the roots of the equation $(a-b+c)x^2+2(a-b)x+(a-b-c)=0$ are rational numbers for all real numbers a and b and for all rational c.

Determine the nature of the roots of the given quadratic equation
${ x }^{ 2 }+x+1=0$

In the given, determine whether the given quadratic equation has real roots and if so, find the roots.
$9{ x }^{ 2 }+7x-2=0$

Number of real roots of ${x} + \sqrt{1- x^2}={5}$

Find the nature of the roots of the following quadratic equations . If the real roots exit. find them:
1) $2{x^2} - 3x + 5 = 0$
2) $3{x^2} - \root 4 \of 3 x + 4 = 0$

Determine the set of values of $k$ for which the given quadratic equation has real roots:
$2x^{2}+kx-4=0$

Show that the equation $x^{2}+ax+1=0$ has real and distinct roots for all real values of $a$ .

Write the nature of the roots of the folliowing quadratic equations
$\quad { 2x }^{ 2 }+6x+3=0$

How many roots of the following quadratic equations are real:
$4x^{2}-4a^{2}x+(a^{4}-b^{4})=0$

$(k+1)x^{2}-2(3k+1)x+8k+1=0$ has equal roots

Find the value of k for which the Q.E $kx^2+1-2(k-1)x+x^2=0$ has equal roots.

Cheek whether the quadratic equation $\left( { a }^{ 2 }+{ b }^{ 2 } \right) { x }^{ 2 }+2\left( ac+bd \right) x+{ c }^{ 2 }+{ d }^{ 2 }=0$ has real roots.

Find the nature of the roots of the following quadratic equation:
1. $3{x^2} + 7x + 2 = 0$
2. ${\left( {x + 3} \right)^2} + 4x + 7 = 0$
3. $2{x^2} - 8x + 34 = 0$
4. ${x^2} + 2\sqrt {3x} - 1 = 0$

Without solving, comment upon the nature of roots of each of the following equations:
$x^{2}-a x-b^{2}=0$

Without solving, comment upon the nature of roots of each of the following equations:
$x^{2}+2 \sqrt{3} x-9=0$

Find the nature of roots of following quadratic equations.
$2x^2+x-1=0$

Write nature of roots by general form of discriminant of equation $ax^2+bx+c=0$ .

Find the value of $k$ , for which following quadratic equations have real and equal roots
$2x^2+kx+3=0$

Find the value of $k$ , for which following quadratic equations have real and equal roots
$x^2-2(k+1)x+k^2=0$

For quadratic equation $ax^2+bx+c=0, a\neq 0$ , at which nature of roots depends?

Describe nature of roots of quadratic equation $ax^2 +bx+c=0, a\neq 0$

Without solving the following quadratic equation, find the greater value of $p$ for which the given equation has real and equal roots. $x^2+(p-3)x+p=0$ .

Prove that the roots of the equation $bx^{2} + (b - c) x + b(b - c - a) = 0$ are real if those of $ax^{2} + 2bx + b = 0$ are imaginary and vice versa.

If $a$ and $c$ are odd prime numbers and $a x^{2}+b x+c=0$ has rational roots where $b \in I .$ Prove that one root of the equation will be independent of $a, b, c$