If $$ax^2 + bx + c = 0$$ then quadratic formula is

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

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

Let $$a, b, c$$ be the lengths of sides of a scalene triangle. If the roots of the equation $$\displaystyle x^{2}+2(a+b+c)x+3\lambda (ab+bc+ca)=0$$ $$(\lambda \in R)$$ are real then

$$\displaystyle \lambda <\dfrac{4}{3}$$

$$\displaystyle \lambda >\frac{5}{3}$$

$$\displaystyle \lambda \in \left ( \frac{1}{3}, \frac{5}{3} \right )$$

$$\displaystyle \lambda \in \left ( \frac{4}{3}, \frac{5}{3} \right )$$

All solutions of the equation $$\displaystyle 4x^2 - 40x + 51 = 0$$ lie in the interval

$$\displaystyle \left(\frac {23}{10}, \frac {83}{10}\right)$$

$$\displaystyle \left(\frac {23}{10}, \frac {15}{2}\right)$$

$$\displaystyle \left(7, \frac {83}{10}\right)$$

The values of $$a$$ for which both the roots of the equation $$(a-6)x^2=a(x-3)$$ are positive and are given by

$$\displaystyle \left( 0,\frac { 72 }{ 11 } \right) $$

$$\displaystyle \left( \frac { 72 }{ 13 } ,6 \right) $$

The quadratic equation $$\displaystyle x^{2}-2x-\lambda=0,\lambda\neq 0,$$

cannot have an integral root if $$n^{2}-1< \lambda< n^{2}+2n$$ where $$n=0,1,2,3,.....$$

cannot have a real root if $$\lambda < -1$$

can have a rational root if $$\lambda$$ is a perfect square

If $$\alpha, \beta$$ are the roots of $$\displaystyle ax^{2}+2bx+c=0$$ and that of $$\displaystyle Ax^{2}+2Bx+C=0$$ be $$\displaystyle \alpha+\delta, \beta +\delta $$ then the value of $$\displaystyle \frac{b^{2}-ac}{B^{2}-AC} $$ is

$$\displaystyle \left ( \frac{a}{A} \right )^{2}$$

$$\displaystyle \left ( \frac{A}{a} \right )^{2}$$

MCQ Questions for Class 10 Maths Quadratic Equations Quiz 10

Which one of the following is the quadratic equation?

0%
$\dfrac{5}{x} - 3 = x^2$
0%
$x(x + 5) = 4$
0%
$n - 1 = 2n$
0%
$\dfrac{1}{x^2} (x + 2) = x$

Explanation

Option : [A]

\frac{5}{x} - 3 = x^{2}

$\dfrac{5}{x} - 3 = x^2$

\frac{5 - 3 x}{x} = x^{2}

$\dfrac{5 - 3x}{x} = x^2$

5 - 3 x = x^{3}

$5 - 3x = x^3$

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

$x^3 + 3x - 5 = 0$

\Rightarrow

$\Rightarrow$ This is not a quadratic equation.

Option : [B]

x (x + 5) = 4

$x (x + 5) = 4$

x^{2} + 5 x = 4

$x^2 + 5x = 4$

x^{2} + 5 x - 4 = 0

$x^2 + 5x - 4 = 0$

\Rightarrow

$\Rightarrow$ This is a quadratic equation.

Option : [C]

n - 1 = 2 n

$n - 1 = 2n$

\Rightarrow

$\Rightarrow$ This is not a quadratic equation.

Option : [D]

\frac{1}{x^{2}} (x + 2) = x

$\dfrac{1}{x^2} (x +2) = x$

x + 2 = x^{3}

$x +2 = x^3$

x^{3} - (x + 2) = 0

$x^3 - (x + 2)=0$

x^{3} - x - 2 = 0

$x^3 - x - 2 = 0$

\Rightarrow

$\Rightarrow$ This is not a quadratic equation.

∴

$\therefore$ Answer is option

[B]

$[B]$

The integral values of

m

$m$ for which the roots of the equation

m x^{2} + (2 m - 1) x + (m - 2) = 0

$m x^{2}+(2 m-1) x+(m-2)=0$ are rational are given by the expression [where

n

$n$ is integer ]

0%
${n}^{2}$
0%
$n(n+2)$
0%
$n(n+1)$
0%
none of these

Explanation

Discriminant

D = (2 m - 1)^{2} - 4 (m - 2) m = 4 m + 1

$D=(2 m-1)^{2}-4(m-2) m=4 m+1$ must be perfect square. Hence,

4 m + 1 = k^{2},

$4 m+1=k^{2},$ say for some

k \in I

$k \in I$

\Rightarrow m = \frac{(k - 1) (k + 1)}{4}

$\Rightarrow \quad m=\dfrac{(k-1)(k+1)}{4}$

Clearly,

k

$k$ must be odd. Let

k = 2 n + 1

$k=2 n+1$

∴ m = \frac{2 n (2 n + 2)}{4} = n (n + 1), n \in I

$\therefore \quad m=\dfrac{2 n(2 n+2)}{4}=n(n+1), n \in I$

Which among the below are quadratic equations?

0%
$\frac{5} {x} - 3 = x^{2}$
0%
$x \left(x + 5\right) = 2$
0%
$n - 1 = 2n$
0%
$\dfrac{1} {x^{2}} \left(x + 2\right) = 2$

a x^{2} + b x + c = 0

$ax^2 + bx + c = 0$ then quadratic formula is

0%
$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
0%
$x=\dfrac{-b\pm\sqrt{b^2+4ac}}{2a}$
0%
$x=\dfrac{-b+ \sqrt{b^2-4ac}}{2a}$
0%
$x=\dfrac{b- \sqrt{b^2-4ac}}{2a}$

Explanation

a x^{2} + b x + c = 0

$ax^2 + bx + c = 0$

quadratic formula

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

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

Discriminant of quadratic equation

3 \sqrt{3} x^{2} + 10 x + \sqrt{3} = 0

$3\sqrt 3x^2+10x+\sqrt 3=0$

0%
$10$
0%
$64$
0%
$46$
0%
$30$

Explanation

Comparing

3 \sqrt{3} x^{2} + 10 x + \sqrt{3}

$3\sqrt 3x^2+10x+\sqrt 3$ by

a x^{2} + b x + c = 0

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

a = \sqrt{3}, b = 10

$a=\sqrt 3, b=10$ and

c = \sqrt{3}

$c=\sqrt 3$
Discriminant

(D) = b^{2} - 4 a c

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

= (10)^{2} - 4 \times 3 \sqrt{3} \times \sqrt{3}

$=(10)^2-4\times 3\sqrt 3 \times \sqrt 3$

= 100 - 4 \times 3 \times 3

$=100-4\times 3\times 3$

= 100 - 36

$=100-36$

= 64

$=64$
Hence, option (B) is correct.

Let

a, b, c

$a, b, c$ be the lengths of sides of a scalene triangle. If the roots of the equation

x^{2} + 2 (a + b + c) x + 3 λ (a b + b c + c a) = 0

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

(λ \in R)

$(\lambda \in R)$ are real then

0%
$\displaystyle \lambda <\dfrac{4}{3}$
0%
$\displaystyle \lambda >\frac{5}{3}$
0%
$\displaystyle \lambda \in \left ( \frac{1}{3}, \frac{5}{3} \right )$
0%
$\displaystyle \lambda \in \left ( \frac{4}{3}, \frac{5}{3} \right )$

Explanation

Since roots of given quadratic are real,

D \geq 0

$D \geq 0$

\Rightarrow (a + b + c)^{2} - 3 λ (a b + b c + c a) \geq 0

$\Rightarrow (a+b+c)^2-3\lambda(ab+bc+ca) \geq 0$

\Rightarrow λ \leq \frac{1}{3} \frac{(a + b + c)^{2}}{a b + b c + c a} = \frac{2}{3} + \frac{1}{3} \frac{a^{2} + b^{2} + c^{2}}{a b + b c + c a}

$\Rightarrow \lambda \leq \cfrac{1}{3} \cfrac{(a+b+c)^2}{ab+bc+ca}=\cfrac{2}{3}+\cfrac{1}{3}\cfrac{a^2+b^2+c^2}{ab+bc+ca}$
Given

a, b, c

$a,b,c$ are sides of triangle

\Rightarrow a + b > c \Rightarrow a c + b c > c^{2}

$\Rightarrow a+b>c\Rightarrow ac+bc>c^2$
Similarly,

a b + b c > b^{2}

$ab+bc> b^2$ and

c a + a b > a^{2}

$ca+ab> a^2$
Using these

a^{2} + b^{2} + c^{2} < 2 (a b + b c + c a)

$a^2+b^2+c^2<2(ab+bc+ca)$
Hence,

λ < \frac{2}{3} + \frac{2}{3} = \frac{4}{3}

$\lambda < \cfrac{2}{3}+\cfrac{2}{3}=\cfrac{4}{3}$ .

The equation

\sqrt{2 x - 2 x^{2} - 5} = x^{2} - 2 x - 3

$\sqrt {2x-2x^2-5}=x^2-2x-3$ , where

x

$x$ is real has

0%
No real solution
0%
Exactly one real solution
0%
Exactly two real solutions
0%
Exactly four real solutions

Explanation

Let us consider the expression,

2 x - 2 x^{2} - 5

$2x-2x^2-5$ .

Δ = 2^{2} - 4 (- 2) (- 5)

$\Delta = 2^2 - 4(-2)(-5)$

Δ = 4 - 40

$\Delta = 4 - 40$

Δ = - 36 < 0

$\Delta = -36 < 0$
The coefficient of

x^{2}

$x^2$ is negative. Hence the expression is negative for all

x

$x$ .
Hence, the term inside the square root is always negative. So, there are no real roots for the equation.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
0%
Assertion is correct but Reason is incorrect.
0%
Assertion is incorrect but Reason is correct.

Explanation

x^{2} + (2 m + 1) x + (2 n + 1) = 0

$x^{ 2 }+(2m+1)x+(2n+1)=0$

D = (2 m + 1)^{2} - 4 (2 n + 1)

$D=(2m+1)^{ 2 }-4(2n+1)$
For

D

$D$ to be a perfect square the expression

4 m^{2} + 4 m + (1 - 8 n - 4)

$4m^2+4m+(1-8n-4)$ which is a quadratic in

m

$m$ must have discriminant zero.

\Rightarrow 4^{2} - (4) (4) (1 - 8 n - 4) = 0

$\Rightarrow 4^2- (4)(4)(1-8n-4) = 0$
But

n

$n$ is an integer.
Therefore, when

m

$m$ &

n

$n$ are integers

D

$D$ cannot be perfect square.
Therefore, roots are irrational.

Ans: A

All solutions of the equation

4 x^{2} - 40 x + 51 = 0

$\displaystyle 4x^2 - 40x + 51 = 0$ lie in the interval

0%
$\displaystyle \left(\frac {23}{10}, \frac {83}{10}\right)$
0%
$\displaystyle \left(\frac {23}{10}, \frac {15}{2}\right)$
0%
$\displaystyle \left(7, \frac {83}{10}\right)$
0%
none of these

The values of

a

$a$ for which both the roots of the equation

(a - 6) x^{2} = a (x - 3)

$(a-6)x^2=a(x-3)$ are positive and are given by

0%
$(0,6)$
0%
$\displaystyle \left( 0,\frac { 72 }{ 11 } \right)$
0%
$\displaystyle \left( 6,\frac { 72 }{ 11 } \right)$
0%
$\displaystyle \left( \frac { 72 }{ 13 } ,6 \right)$

Explanation

The given equation is

(a - 6) x^{2} - a x + 3 a = 0

$(a-6){x}^{2}-ax+3a=0$

D = a^{2} - 4 (a - 6) .3 a > 0 \Rightarrow 0 < a < \frac{72}{11}

$\displaystyle D={ a }^{ 2 }-4\left( a-6 \right) .3a>0\Rightarrow 0<a<\frac { 72 }{ 11 }$ ...(i)

Since, both the roots are positive, so sum & product of roots are positive

∴

$\therefore$

α + β = \frac{a}{a - 6} > 0 \Rightarrow a < 0

$\displaystyle \alpha +\beta =\frac { a }{ a-6 } >0\Rightarrow a<0$ or

a > 6

$a>6$ ...(ii)

and

α β = \frac{3 a}{a - 6} > 0 \Rightarrow a < 0

$\displaystyle \alpha \beta =\frac { 3a }{ a-6 } >0\Rightarrow a<0$ or

a > 6

$a>6$

∴ 6 < a < \frac{72}{11}

$\displaystyle \therefore 6<a<\frac { 72 }{ 11 }$

[

$[$ From Eqs.(i) and (ii)

]

$]$

a =

(6, \frac{72}{11})

$\displaystyle \left( 6,\frac { 72 }{ 11 } \right)$

The roots of

(x - a) (x - c) + k (x - b) (x - d) = 0

$(x-a)(x-c)+k(x-b)(x-d)=0$ are real and distinct for all real

k

$k$ if

0%
$a < b < c < d$
0%
$a > b < c < d$
0%
$a < b > c < d$
0%
$a < b = c < d$

Explanation

The roots of $(x-a)(x-c)+k(x-b)(x-d)=0$ are real and distinct for all real $k$

$f(x)=(1+k){ x }^{ 2 }-(a+c+k(b+d))x+ac+kbd$

As $D>0$ for real roots.

So ,

${(a+c+k(b+d)) }^{ 2}-4(1+k)(ac+kbd)>0$

$\Rightarrow$ $(b+d)^{ 2 }{ k }^{ 2}+2(a+c)(b+d)k+(a+c)^{ 2 }-4(bd{ k }^{ 2 }+(ac+bd)k+ac)>0$

$\Rightarrow$ $(b-d)^{ 2 }{ k }^{ 2}+(2(a+c)(b+d)-4(ac+bd))k+(a-c)^{ 2 }>0$

This true for all $k$

$D<0$

$(2(a+c)(b+d)-4(ac+bd))^{ 2 }-4((a-c)(b-d))^{ 2 }<0$

$((a+c)(b+d)-2(ac+bd))^{ 2 }-((a-c)(b-d))^{ 2 }<0$

Applying ${a }^{ 2 }-b^{ 2 }=(a-b)(a+b)$ we get,

$\Rightarrow$ $((a+c)(b+d)-2(ac+bd)+(a-c)(b-d))((a+c)(b+d)-2(ac+bd)-(a-c)(b-d))<0$

$\Rightarrow$ $(ab+ad+bc+cd-2ac-2bd+ab-ad-bc+cd)(ab+ad+bc+cd-2ac-2bd-ab+ad+bc-cd)<0$

$(ab+cd-ac-bd)(ad+bc-ac-bd)<0$

$(a-d)(b-c)(a-b)(d-c)<0$

$\Rightarrow$ $a<b<c<d$

The quadratic equation

x^{2} - 2 x - λ = 0, λ \neq 0,

$\displaystyle x^{2}-2x-\lambda=0,\lambda\neq 0,$

0%
cannot have a real root if $\lambda < -1$
0%
can have a rational root if $\lambda$ is a perfect square
0%
cannot have an integral root if $n^{2}-1< \lambda< n^{2}+2n$ where $n=0,1,2,3,.....$
0%
none of these

Explanation

x^{2} - 2 x - λ = 0

$x^{ 2 }-2x-\lambda =0$

D = 4 + 4 λ < 0

$D=4+4\lambda <0$
Therefore, equation has no real root when

λ < - 1

$\lambda<-1$ .
and cannot have an integral root if

n^{2} < λ + 1

$n^{ 2 }<\lambda+1$ .

1

$1$ is a root of the equations

a y^{2}

$ay^2$ +

a y + 3 = 0

$ay + 3 = 0$ and

y^{2}

$y^2$ +

y + b = 0

$y + b = 0$ , then find the value of

a b

$ab$

0%
$3$
0%
$4$
0%
$5$
0%
$6$

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

${ x }^{ 2 }-\left( a+b+c \right) x+\left( ab+bc+ca \right) =0$ has non real roots, where

a, b, c \in R^{+}

$a,b,c\in { R }^{ + }$ , then

\sqrt{a}, \sqrt{b}, \sqrt{c}

$\sqrt { a } ,\sqrt { b } ,\sqrt { c }$

0%
Can be the sides of a triangle
0%
Cannot be the sides of triangle
0%
Nothing can be said
0%
None of these

Explanation

b^{2} - 4 a c < 0

$b^{2}-4ac<0$
Or

(a + b + c)^{2} - 4 (a b + b c + a c) < 0

$(a+b+c)^{2}-4(ab+bc+ac)<0$

a^{2} + b^{2} + c^{2} - 2 a b - 2 b c - 2 a c < 0

$a^{2}+b^{2}+c^{2}-2ab-2bc-2ac<0$

(a - b - c)^{2} - 4 b c < 0

$(a-b-c)^{2}-4bc<0$
or

(a - b - c)^{2} < 4 b c

$(a-b-c)^{2}<4bc$

(a - b - c) < 2 \sqrt{b} \sqrt{c}

$(a-b-c)<2\sqrt{b}\sqrt{c}$

a < b + c + 2 \sqrt{b} \sqrt{c}

$a<b+c+2\sqrt{b}\sqrt{c}$

a < (\sqrt{b} + \sqrt{c})^{2}

$a<(\sqrt{b}+\sqrt{c})^{2}$

Or

{\sqrt{a}}^{2} < (\sqrt{b} + \sqrt{c})^{2}

$\sqrt{a}^{2}<(\sqrt{b}+\sqrt{c})^{2}$

\sqrt{a} < \sqrt{b} + \sqrt{c}

$\sqrt{a}<\sqrt{b}+\sqrt{c}$
Hence sum of two sides is greater than the third side.
Hence a triangle can be formed with sides

\sqrt{a}, \sqrt{b}, \sqrt{c}

$\sqrt{a},\sqrt{b},\sqrt{c}$ .

The value of

x^{2} - 6 x + 13

$x^2-6x+13$ can never be less than

0%
$4$
0%
$5$
0%
$4.5$
0%
$7$

Explanation

∵ x^{2} - 6 x + 13 = x^{2} - 6 x + 9 + 4

$\because x^2-6x+13=x^2-6x+9+4$

= (x - 3)^{2} + 4 \geq 4

$=(x-3)^2+4\ge4$
because

(x - 3)^{2}

$(x-3)^2$ can never be less than zero

∴

$\therefore$ least value of

x^{2} - 6 x + 13 = 4

$x^2-6x+13=4$

The number of values of

k

$\displaystyle k$ for which

(x^{2} - (k - 2) x + k^{2}) (x^{2} + k x + (2 k - 1))

$\displaystyle \left ( x^{2} - \left ( k - 2 \right )x + k^{2} \right ) \left ( x^{2} + kx + \left ( 2k - 1 \right ) \right )$ is a perfect square

0%
$\displaystyle 1$
0%
$\displaystyle 2$
0%
$\displaystyle 3$
0%
$\displaystyle 0$

Explanation

Given expression

(x^{2} - (k - 2) x + k^{2}) (x^{2} + k x + (2 k - 1))

$\displaystyle \left ( x^{2} - \left ( k - 2 \right )x + k^{2} \right ) \left ( x^{2} + kx + \left ( 2k - 1 \right ) \right )$
For the expression to be a perfect square, the two quadratic factors should have coefficients of

x

$x$ and constant terms equal.

\Rightarrow - (k - 2) = k

$\Rightarrow -(k-2)=k$ and

k^{2} = 2 k - 1

$k^2=2k-1$

\Rightarrow k = 1

$\Rightarrow k=1$ and

(k - 1)^{2} = 0

$(k-1)^2=0$

\Rightarrow k = 1

$\Rightarrow k=1$
Hence, the number of values of

k

$k$ is 1.

Find the root(s) of the equation

y + \sqrt{y + 5} = 7

$y+\sqrt{y+5}=7$

0%
11
0%
4
0%
4 and 11
0%
$\pm 4$
0%
none of these

Explanation

Given

y + \sqrt{y + 5} = 7

$y+\sqrt{y+5} = 7$

\Rightarrow y - 7 = \sqrt{y + 5}

$\Rightarrow y-7 = \sqrt{y+5}$
Now by squaring on both sides, we get

{(y - 7)}^{2} = y + 5

${(y-7)}^{2}=y+5$

\Rightarrow y^{2} - 15 y + 44 = 0

$\Rightarrow {y}^{2}-15y+44=0$
Therefore we get

y = \frac{(15 \pm 7)}{2} = 11, 4

$y = \dfrac {(15 \pm 7)}{2} = 11 , 4$
Now substitute

y = 4

$y=4$ in given equation, we get

4 + 3 = 7

$4+3=7$ (TRUE)
Now substitute

y = 11

$y=11$ in given equation, we get

11 + 4 = 7

$11+4=7$ (FALSE)
Therefore

y = 4

$y=4$ is the only solution of given equation

α, β

$\alpha, \beta$ are the roots of

a x^{2} + 2 b x + c = 0

$\displaystyle ax^{2}+2bx+c=0$ and that of

A x^{2} + 2 B x + C = 0

$\displaystyle Ax^{2}+2Bx+C=0$ be

α + δ, β + δ

$\displaystyle \alpha+\delta, \beta +\delta$ then the value of

\frac{b^{2} - a c}{B^{2} - A C}

$\displaystyle \frac{b^{2}-ac}{B^{2}-AC}$ is

0%
$\displaystyle \left ( \frac{a}{A} \right )^{2}$
0%
$\displaystyle \left ( \frac{A}{a} \right )^{2}$
0%
$0$
0%
$1$

Explanation

\Rightarrow

$\Rightarrow$

α

$\alpha$ and

β

$\beta$ are the roots of quadratic equation

a x^{2} + 2 b x + c = 0

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

\Rightarrow

$\Rightarrow$

α β = \frac{c}{a}

$\alpha\beta=\dfrac{c}{a}$ ----- ( 1 )

\Rightarrow

$\Rightarrow$

α + β = \frac{- 2 b}{a}

$\alpha+\beta=\dfrac{-2b}{a}$

\Rightarrow

$\Rightarrow$

(α + β)^{2} = α^{2} + β^{2} + 2 α β

$(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta$

\Rightarrow

$\Rightarrow$

{(\frac{- 2 b}{a})}^{2} = α^{2} + β^{2} + 2 \times \frac{c}{a}

$\left(\dfrac{-2b}{a}\right)^2=\alpha^2+\beta^2+2\times\dfrac{c}{a}$

∴

$\therefore$

α^{2} + β^{2} = \frac{4 b^{2}}{a^{2}} - \frac{2 c}{a}

$\alpha^2+\beta^2=\dfrac{4b^2}{a^2}-\dfrac{2c}{a}$ ----- ( 2 )

\Rightarrow

$\Rightarrow$

(α - β)^{2} = α^{2} + β^{2} - 2 α β

$(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta$

\Rightarrow

$\Rightarrow$

(α - β)^{2} = \frac{4 b^{2}}{a^{2}} - \frac{2 c}{a} - \frac{2 c}{a}

$(\alpha-\beta)^2=\dfrac{4b^2}{a^2}-\dfrac{2c}{a}-\dfrac{2c}{a}$ [ From ( 1 ) and ( 2 ) ]

\Rightarrow

$\Rightarrow$

(α - β)^{2} = \frac{4 b^{2} - 4 a c}{a^{2}}

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

∴

$\therefore$

α - β = \sqrt{\frac{4 b^{2} - 4 a c}{a^{2}}}

$\alpha-\beta=\sqrt{\dfrac{4b^2-4ac}{a^2}}$ ------ ( 3 )

\Rightarrow

$\Rightarrow$ Now,

α + δ

$\alpha+\delta$ and

β + δ

$\beta+\delta$ are roots of quadratic equation

A x^{2} + 2 B x + C = 0

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

\Rightarrow

$\Rightarrow$

(α + δ) (β + δ) = \frac{C}{A}

$(\alpha+\delta)(\beta+\delta)=\dfrac{C}{A}$

\Rightarrow

$\Rightarrow$

α β + α δ + β δ + δ^{2} = \frac{C}{A}

$\alpha\beta+\alpha\delta+\beta\delta+\delta^2=\dfrac{C}{A}$

∴

$\therefore$

α β = \frac{C}{A} - α δ - β δ - δ^{2}

$\alpha\beta=\dfrac{C}{A}-\alpha\delta-\beta\delta-\delta^2$ -------- ( 4 )

\Rightarrow

$\Rightarrow$

(α + δ) + (β + δ) = \frac{- 2 B}{A}

$(\alpha+\delta)+(\beta+\delta)=\dfrac{-2B}{A}$

\Rightarrow

$\Rightarrow$

α + β + 2 δ = \frac{- 2 B}{A}

$\alpha+\beta+2\delta=\dfrac{-2B}{A}$

\Rightarrow

$\Rightarrow$

α + β = \frac{- 2 B}{A} - 2 δ

$\alpha+\beta=\dfrac{-2B}{A}-2\delta$ -------- ( 5 )

\Rightarrow

$\Rightarrow$

(α + β)^{2} = {(\frac{- 2 B}{A} - 2 δ)}^{2}

$(\alpha+\beta)^2=\left(\dfrac{-2B}{A}-2\delta\right)^2$

∴

$\therefore$

α^{2} + β^{2} = {(\frac{- 2 B}{A} - 2 δ)}^{2} - 2 α β

$\alpha^2+\beta^2=\left(\dfrac{-2B}{A}-2\delta\right)^2-2\alpha\beta$ -------- ( 6 )

\Rightarrow

$\Rightarrow$

[(α + δ) - (β + δ)]^{2} = (α - β)^{2}

$[(\alpha+\delta)-(\beta+\delta)]^2=(\alpha-\beta)^2$

\Rightarrow

$\Rightarrow$

(α - β)^{2} = α^{2} + β^{2} - 2 α β

$(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta$

\Rightarrow

$\Rightarrow$

(α - β)^{2} = {(\frac{- 2 B}{A} - 2 δ)}^{2} - 2 α β - 2 α β

$(\alpha-\beta)^2=\left(\dfrac{-2B}{A}-2\delta\right)^2-2\alpha\beta-2\alpha\beta$ [ From ( 6 ) ]

\Rightarrow

$\Rightarrow$

(α - β)^{2} = {(\frac{- 2 B}{A} - 2 δ)}^{2} - 4 α β

$(\alpha-\beta)^2=\left(\dfrac{-2B}{A}-2\delta\right)^2-4\alpha\beta$

\Rightarrow

$\Rightarrow$

(α - β)^{2} = \frac{4 B^{2}}{A^{2}} + \frac{8 B δ}{A} + 4 δ^{2} - \frac{4 C}{A} + 4 α δ + 4 β δ + 4 δ^{2}

$(\alpha-\beta)^2=\dfrac{4B^2}{A^2}+\dfrac{8B\delta}{A}+4\delta^2-\dfrac{4C}{A}+4\alpha\delta+4\beta\delta+4\delta^2$ [ From ( 4 ) ]

\Rightarrow

$\Rightarrow$

(α - β)^{2} = \frac{4 B}{A^{2}} - \frac{4 C}{A} + \frac{8 B δ}{A} + 8 δ^{2} + 4 δ (α + β)

$(\alpha-\beta)^2=\dfrac{4B}{A^2}-\dfrac{4C}{A}+\dfrac{8B\delta}{A}+8\delta^2+4\delta(\alpha+\beta)$

\Rightarrow

$\Rightarrow$

(α - β)^{2} = \frac{4 B^{2} - 4 A C}{A^{2}} + \frac{8 B δ}{A} + 8 δ^{2} - \frac{8 B δ}{A} - 8 δ^{2}

$(\alpha-\beta)^2=\dfrac{4B^2-4AC}{A^2}+\dfrac{8B\delta}{A}+8\delta^2-\dfrac{8B\delta}{A}-8\delta^2$ [ From ( 5 ) ]

\Rightarrow

$\Rightarrow$

(α - β)^{2} = \frac{4 B^{2} - 4 A C}{A^{2}}

$(\alpha-\beta)^2=\dfrac{4B^2-4AC}{A^2}$

∴

$\therefore$

(α - β) = \sqrt{\frac{4 B^{2} - 4 A C}{A^{2}}}

$(\alpha-\beta)=\sqrt{\dfrac{4B^2-4AC}{A^2}}$ ------- ( 7 )

\Rightarrow

$\Rightarrow$

\frac{α - β}{α - β} = \frac{\sqrt{\frac{4 b^{2} - 4 a c}{a^{2}}}}{\sqrt{\frac{4 B^{2} - 4 A C}{A^{2}}}}

$\dfrac{\alpha-\beta}{\alpha-\beta}=\dfrac{\sqrt{\dfrac{4b^2-4ac}{a^2}}}{\sqrt{\dfrac{4B^2-4AC}{A^2}}}$ [ Dividing ( 3 ) by ( 7 )]

\Rightarrow

$\Rightarrow$

1 = \frac{\sqrt{\frac{4 b^{2} - 4 a c}{a^{2}}}}{\sqrt{\frac{4 B^{2} - 4 A C}{A^{2}}}}

$1=\dfrac{\sqrt{\dfrac{4b^2-4ac}{a^2}}}{\sqrt{\dfrac{4B^2-4AC}{A^2}}}$

\Rightarrow

$\Rightarrow$

\frac{4 B^{2} - 4 A C}{A^{2}} = \frac{4 b^{2} - 4 a c}{a^{2}}

$\dfrac{4B^2-4AC}{A^2}=\dfrac{4b^2-4ac}{a^2}$

∴

$\therefore$

\frac{b^{2} - a c}{B^{2} - A C} = \frac{a^{2}}{A^{2}} = {(\frac{a}{A})}^{2}

$\dfrac{b^2-ac}{B^2-AC}=\dfrac{a^2}{A^2}=\left(\dfrac{a}{A}\right)^2$

The roots of the equation

(x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a) = 0

$\displaystyle \left ( x-a \right )\left ( x-b \right )+\left ( x-b \right )\left ( x-c \right )+\left ( x-c \right )\left ( x-a \right )=0$ are

0%
positive
0%
negative
0%
real
0%
imaginary

Explanation

Given equation is written as

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

$\displaystyle 3x^{2}-2x\left ( a+b+c \right )+\left ( ab+bc+ca \right )=0$
Discriminant is

4 {(a + b + c)}^{2} - 12 (a b + b c + c a)

$\displaystyle 4\left ( a+b+c \right )^{2}-12\left ( ab+bc+ca \right )$

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

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

= 2 [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]

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

∴

$\displaystyle \therefore$ Roots are real

If the quadratic equation

x^{2} - m x - 4 x + 1 = 0

$\displaystyle x^{2}-mx-4x+1=0$ has real and distinct roots, then the values of

m

$m$ are
A.

(- \infty, - 6)

$\displaystyle (-\infty , -6)$
B.

(- \infty, - 3)

$\displaystyle (-\infty,-3)$

(- 2, \infty)

$(-2,\infty )$

(2, \infty)

$\displaystyle \left ( 2,\infty \right )$

0%
A or C
0%
A or D
0%
B or C
0%
B or D

Explanation

For an equation

a x^{2} + b x + c = 0

$a{x}^{2} + bx + c = 0$ , the discriminant

△ = b^{2} - 4 a c

$\triangle = {b}^{2} -4ac$ helps us understand the nature of the roots.

When

△ > 0

$\triangle > 0$ and a perfect square, roots are real and distinct

So

{(- (m + 4))}^{2} - 4 \times 1 \times 1 > 0

${(-(m+4))}^{2} -4 \times 1 \times 1 > 0$

\Rightarrow

$\Rightarrow$

(m^{2} + 16 + 8 m) - 4 > 0

$({m}^{2} + 16 +8m) - 4 > 0$

\Rightarrow

$\Rightarrow$

m^{2} + 8 m + 12 > 0

${m}^{2} +8m + 12 > 0$

\Rightarrow

$\Rightarrow$

m^{2} + 2 m + 6 m + 12 > 0

${m}^{2} +2m + 6m + 12 > 0$

\Rightarrow

$\Rightarrow$

m (m + 2) + 6 (m + 2) > 0

$m(m+2) + 6(m+2) > 0$

\Rightarrow

$\Rightarrow$

(m + 6) (m + 2) > 0

$(m+6)(m+2) > 0$

When both

(m + 6) > 0; (m + 2) > 0

$(m+6) > 0 ; (m+2) > 0$

\Rightarrow

$\Rightarrow$

m > - 6; m > - 2

$m > -6 ; m > -2$

From these two solutions, we have

m > - 2

$m > -2$

Also,

(m + 6) (m + 2) > 0

$(m+6)(m+2) > 0$ , when both

(m + 6) < 0; (m + 2) < 0

$(m+6) < 0 ; (m+2) < 0$

\Rightarrow

$\Rightarrow$

m < - 6; m < - 2

$m < -6 ; m< -2$

From these two solutions, we have

m < - 6

$m <- 6$

So both

A

$A$ and

C

$C$ are the possible values of

m

$m$

Let

f (x) = x^{2} - 3 x + 4

$\displaystyle f(x)= x^{2}-3x+4$ , then the value of

x

$x$ which satisfies

f (1) + f (x) = f (1) f (x)

$\displaystyle f(1)+f(x)= f(1)f(x)$ is

0%
$1$
0%
$2$
0%
$1 \ or\ 2$
0%
$1\ and\ 0$

Explanation

f (x) = x^{2} - 3 x + 4 = 0

$\displaystyle f(x)= x^{2}-3x+4= 0$

∴ f (1) = 2

$\displaystyle \therefore f(1)= 2$ and

x

$x$ satisfies the condition

f (1) + f (x) = f (1) f (x)

$\displaystyle f(1)+f(x)= f(1)f(x)$

∴ 2 + x^{2} - 3 x + 4 = 2 (x^{2} - 3 x + 4)

$\displaystyle \therefore 2+x^{2}-3x +4= 2(x^{2}-3x+4)$

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

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

The equation

9 y^{2} (m + 3) + 6 (m - 3) y + (m + 3) = 0

$\displaystyle 9y^{2}(m+3)+6(m-3)y+(m+3)=0$ , where

m

$m$ is real has real roots then

0%
$\displaystyle m< 0$
0%
$\displaystyle m> 0$
0%
$\displaystyle m\leq 0$
0%
$\displaystyle m\geq 0$

Explanation

For an equation

a x^{2} + b x + c = 0

$a{x}^{2} + bx + c = 0$ , the discriminant

△ = b^{2} - 4 a c

$\triangle = {b}^{2} -4ac$ helps us understand the nature of the roots.

When

△ \geq 0

$\triangle \ge 0$ then roots are real and equal.

{[6 (m - 3)]}^{2} - 4 \times 9 (m + 3) \times (m + 3) \geq 0

${[6(m-3)]}^{2} -4 \times 9(m+3) \times (m + 3) \ge 0$

36 (m^{2} + 9 + - 6 m) - 36 (m^{2} + 9 + 6 m) \geq 0

$36({m}^{2} + 9 +-6m) - 36( {m}^{2} + 9 + 6m) \ge 0$

36 m^{2} + 324 - 216 m - 36 m^{2} - 324 - 216 m \geq 0

$36{m}^{2} + 324 - 216m - 36{m}^{2} -324 -216m \ge 0$

- 432 m \geq 0

$-432m \ge 0$

∴ m \leq 0

$\therefore m \le 0$

x^{2} + (a - b) x + (1 - a - b) = 0,

$\displaystyle x^{2}+(a-b)x+(1-a-b)= 0,$ where

a, b \in R,

$\displaystyle a,b \in R,$ the value of

a

$a$ such that the equation has distinct real roots for all value of

b

$b$ are

0%
$\displaystyle a> 1$
0%
$\displaystyle a< 1$
0%
$\displaystyle a> 2$
0%
$\displaystyle a< 2$

Explanation

For real and distinct roots

Δ > 0

$\Delta >0$

\Rightarrow

$\Rightarrow$

(a - b)^{2} - 4 (1 - a - b) > 0

$(a-b)^{2}-4(1-a-b)>0$

\Rightarrow

$\Rightarrow$

(a - b - 2)^{2} + 8 a - 8 > 0

$(a-b-2)^{2} +8a-8>0$

\Rightarrow

$\Rightarrow$

8 a - 8 > 0

$8a-8>0$

a > 1

$a>1$

A ray emanating from (6,2) is incident on ellipse

\frac{(x - 1)^{2}}{45} + \frac{(y - 2)^{2}}{20} = 1

$\displaystyle \frac{(x-1)^{2}}{45}+\frac{(y-2)^{2}}{20}=1$ at (4.6) The equation of reflected ray (after 1st reflection ) is

0%
x - 2y + 8 =0
0%
x + 2y + 8 =0
0%
x + 2y - 8 = 0
0%
x - 2y - 8 =0

The number of real roots of the equation

(x - 1)^{2} + (x - 2)^{2} + (x - 3)^{2} = 0

$(x - 1)^2 + (x - 2)^2 + (x- 3)^2 = 0$ is :

0%
$1$
0%
$2$
0%
$3$
0%
None of these

Explanation

Given,

(x - 1)^{2} + (x - 2)^{2} + (x - 3)^{2} = 0

$(x - 1)^2 + (x - 2)^2 + (x - 3)^2 = 0$

\Rightarrow x^{2} - 2 x + 1 + x^{2} - 4 x + 4 + x^{2} - 6 x + 9 = 0

$\Rightarrow x^2 - 2x + 1 + x^2 - 4x + 4 + x^2 - 6x + 9 = 0$

\Rightarrow 3 x^{2} - 12 x + 14 = 0

$\Rightarrow 3x^2 - 12x + 14 = 0$

Let

D

$D$ be the discriminant of the obtained quadratic equation.

D = (- 12)^{2} - 4 \times 14 \times 3 = 144 - 168 = - 24

$D=(-12)^2-4\times 14\times 3 \\=144-168\\=-24$

Since the value of discriminant is negative, the equation does not have any real roots

Given that

r

$r$ and

s

$s$ are constants , the solution of the quadratic equation,

r x^{2} = \frac{1}{s} x + 3

$r{x}^{2}=\dfrac{1}{s}x+3$ , is:

0%
$x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } +12r } }{ 2r }$
0%
$x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { -\dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2sr }$
0%
$x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2r }$
0%
$x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { { s }^{ 2 }-12sr } }{ 2sr }$

Explanation

Given, $rx^{2}=\dfrac{1}{s}x+3$

$\Rightarrow rx^{2}-\dfrac{1}{s}x-3=0$

WE know that in equation $ax^{2}+bx+c=0$

The roots of equation

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

In given equation $rx^{2}-\dfrac{1}{s}x-3=0$

compare with above equation we get $a=r$ , $b=-\dfrac{1}{s}$ and $c=-3$

Then $x=\dfrac{-\dfrac{-1}{s}\pm \sqrt{\left (- \dfrac{1}{s} \right )^{2}-4\times r\times -3}}{2r}$

$x=\dfrac{\dfrac{1}{s}\pm \sqrt{\dfrac{1}{s^{2}+12r}}}{2r}$

$\Rightarrow x=\dfrac{1}{2sr}\pm\dfrac{\sqrt{\dfrac{1}{s^{2}}+12r}}{2r}$

If the expression

x^{2} + 2 (a + b + c) x + 3 (b c + c a + a b)

$x^2+2(a+b+c)x+3(bc+ca+ab)$ is a perfect square, then

0%
$a=b=c$
0%
$a=\pm b=\pm c$
0%
$a=b\neq c$
0%
None of the above

Explanation

x^{2} + 2 (a + b + c) x + 3 (a b + b c + a c) i s a p e r f e c t s q u a r e ∴ D = 0 4 {(a + b + c)}^{2} - 4.3 (a b + b c + a c) = 0 a^{2} + b^{2} + c^{2} + 2 (a b + b c + a c) - 3 (a b + b c + a c) = 0 a^{2} + b^{2} + c^{2} - a b - b c - a c = 0 ∴ \frac{1}{2} [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}] = 0 ∴ a = b = c A s s u m o f s q u a r e s i s z e r o o n l y i f s q u a r e a r e z e r o

${ x }^{ 2 }+2(a+b+c)x+3(ab+bc+ac)\quad is\quad a\quad perfect\quad square\\ \therefore D=0\\ 4{ (a+b+c) }^{ 2 }-4.3(ab+bc+ac)=0\\ { a }^{ 2 }+b^{ 2 }+{ c }^{ 2 }+2(ab+bc+ac)-3(ab+bc+ac)=0\\ { a }^{ 2 }+b^{ 2 }+{ c }^{ 2 }-ab-bc-ac=0\\ \therefore \cfrac { 1 }{ 2 } [{ (a-b) }^{ 2 }+{ (b-c) }^{ 2 }+{ (c-a) }^{ 2 }]=0\\ \therefore a=b=c\\ As\quad sum\quad of\quad squares\quad is\quad zero\quad only\quad if\quad square\quad are\quad zero$

Number of possible value(s) of integer '

a

$a$ ' for which the quadratic equation

x^{2} + a x + 16 = 0

$x^2+ax+16=0$ has equal roots, is

0%
$4$
0%
$6$
0%
$2$
0%
none of these

Explanation

Given the quadratic equation

x^{2} + a x + 16 = 0

$x^2+ax+16=0$

= a^{2} - 4 \times 16 = 0

$=a^2 -4 \times 16 =0$

⟹ a = + 8, - 8

$\Longrightarrow a=+8,-8$

If, for a positive integer n, the quadratic equation,

x (x + 1) + (x + 1) (x + 2) + \dots + (x + \frac{1}{n - 1}) (x + n) = 10 n

$x ( x + 1 ) + ( x + 1 ) ( x + 2 ) + \ldots + \left( x + \frac { 1 } { n - 1 } \right) ( x + n ) = 10 n$ has two consecutive integral solutions, then n is equal to

0%
11
0%
12
0%
9
0%
10

Explanation

n x^{2} + x (1 + 3 + 5 + . . . + (2 n - 1)) + (1.2 + 2.3 + . . . + (n - 1) . n) - 10 n = 0

$n{x}^{2}+x\left(1+3+5+...+\left(2n-1\right)\right)+\left(1.2+2.3+...+\left(n-1\right).n\right)-10n=0$

\Rightarrow n x^{2} + x n^{2} + \frac{n (n^{2} - 1)}{3} - 10 n = 0

$\Rightarrow n{x}^{2}+x{n}^{2}+\dfrac{n\left({n}^{2}-1\right)}{3}-10n=0$

\Rightarrow x^{2} + n x + \frac{(n^{2} - 1)}{3} - 10 = 0

$\Rightarrow {x}^{2}+nx+\dfrac{\left({n}^{2}-1\right)}{3}-10=0$

Let

α

$\alpha$ and

β

$\beta$ be the roots of the equation then

α + β = - n

$\alpha+\beta=-n$ and

α β = \frac{(n^{2} - 1)}{3} - 10

$\alpha\beta=\dfrac{\left({n}^{2}-1\right)}{3}-10$

And

{(α - β)}^{2} = 1

${\left(\alpha-\beta\right)}^{2}=1$

\Rightarrow {(α + β)}^{2} - 4 α β = 1

$\Rightarrow {\left(\alpha+\beta\right)}^{2}-4\alpha\beta=1$

\Rightarrow n^{2} - 4 (\frac{(n^{2} - 1)}{3} - 10) = 1

$\Rightarrow {n}^{2}-4\left(\dfrac{\left({n}^{2}-1\right)}{3}-10\right)=1$

\Rightarrow 3 n^{2} - 4 n^{2} + 4 + 120 - 3 = 0

$\Rightarrow 3{n}^{2}-4{n}^{2}+4+120-3=0$

\Rightarrow n^{2} = 121

$\Rightarrow {n}^{2}=121$

∴ n = 11

$\therefore n=11$

If the root of the equation

x^{2} + p x + q = 0

$x^{2}+px+q=0$ differ from the roots of the equation

x^{2} + q x + p = 0

$x^{2}+qx+p=0$ by the same quantity then

0%
$p+q+1=0$
0%
$p+q+2=0$
0%
$p+q+4=0$
0%
None of these

Explanation

We have,

${{x}^{2}}+px+q=0\,......\,\left( 1 \right)$

${{x}^{2}}+qx+p=0\,.......\,\left( 2 \right)$

By equation (1)

Let $a$ and $b$ be the roots of equation (1)

Then,

Sum of roots $a+b=\dfrac{-p}{1}=-p$

Product $a.b=\dfrac{q}{1}=q$

According to given question,

Difference of roots

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

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

$a-b=\sqrt{{{p}^{2}}-4q}$

Now, by equation (2)

Let $c$ and $d$ be the roots of equation (2)

Then,

Sum of roots $c+d=\dfrac{-q}{1}=-q$

Product $a.b=\dfrac{p}{1}=p$

According to given question,

Difference of roots

$c-d=\sqrt{{{\left( c+d \right)}^{2}}-4cd}$

$c-d=\sqrt{{{\left( -p \right)}^{2}}-4q}$

$c-d=\sqrt{{{p}^{2}}-4q}$

Again, according to given question,

Difference of roots of both equations are equal

Therefore,

$a-b=c-d$

$\sqrt{{{q}^{2}}-4p}=\sqrt{{{p}^{2}}-4q}$

${{q}^{2}}-4p={{p}^{2}}-4q$

${{q}^{2}}-{{p}^{2}}=4p-4q$

$\left( q-p \right)\left( q+p \right)=4\left( p-q \right)$

$\left( q-p \right)\left( p+q \right)=-4$

$\left( q-p \right)\left( p+q+4 \right)=0$

$q=p\,and\,p+q+4=0$

Hence, this is the answer.

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 10 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 10 - MCQExams.com

0:0:3

Practice Class 10 Maths Quiz Questions and Answers

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