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

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

$$x^{2}+x+1=0$$
In $$ax^{2}+bx+c=0$$ 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,
$$ x=\dfrac { -b\pm \sqrt { { b }^{ 2 }-4ac }  }{ 2a }$$

As $$b^2 - 4ac \geq 0$$, 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.
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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

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

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

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

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

If $$D=0$$, then

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

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

or   $$a=b=c$$


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

We know that while finding the root of a quadratic equation $$ax^2+bx+c=0$$ by quadratic formula $$x=\frac { -b±\sqrt { b^{ 2 }-4ac }  }{ 2a }$$, 
if $$b^2-4ac>0$$, then the roots are real and distinct 
if $$b^2-4ac=0$$, then the roots are real and equal and
if $$b^2-4ac<0$$, 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,
$$D=b^2-4ac$$
     $$ = {\left( { - 4{a^2}} \right)^2} - 4\left( 4 \right)\left( {{a^4} - {b^4}} \right)$$
     $$ = 16{a^4} - 16\left( {{a^4} - {b^4}} \right)$$
     $$ = 16{a^4} - 16{a^4} + 16{b^4}$$
     $$ = 16{b^4}$$

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


For what value/values of $$k$$:
 $$(k+1)x^{2}-2(3k+1)x+8k+1=0$$ 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.$${x^2} + 2\sqrt {3x} - 1 = 0$$

A. $$3{ x }^{ 2 }+7x+2=0$$
$$\triangle ={ b }^{ 2 }-4ac={ 7 }^{ 2 }-4\left( 2 \right) \left( 3 \right) =49-24=25>0$$
$$\therefore $$Roots are real and positive
B. $${ \left( x+3 \right)  }^{ 2 }+4x+7=0$$
$${ x }^{ 2 }+9+6x+4x+7=0$$
$${ x }^{ 2 }+10x+16=0$$
$$\triangle ={ b }^{ 2 }-4ac=100-4\left( 16 \right) =36={ 6 }^{ 2 }>0$$
$$\therefore $$Roots are real and positive
C. $$2{ x }^{ 2 }-8x+34=0$$
$$\triangle ={ 8 }^{ 2 }-4\left( 2 \right) \left( 34 \right) <0$$
$$\therefore $$ Roots are imaginary
D. $${ x }^{ 2 }+2\sqrt { 3 } x-1=0$$
$$\triangle ={ \left( 2\sqrt { 3 }  \right)  }^{ 2 }+4\left( 1 \right) =12+4=16>0$$
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
$${ \left( b-c \right)  }^{ 2 }-4.b.\left( b-c-a \right) >0$$
Or
$$\left( { b }^{ 2 }+{ c }^{ 2 }-2bc \right) -4\left( { b }^{ 2 }-bc-ab \right) >0$$
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
$$4{ b }^{ 2 }-4ab<0\\ 4\left( { b }^{ 2 }-ab \right) <0\\ 4b\left( b-a \right) <0$$
Or
$$\left( a-b \right) >0$$





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 $$ a, b, c $$


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Class 11 Commerce Maths Extra Questions