Quadratic Equations Test

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Quadratic Equations Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

If $1$ is a root of the equations $ay^2$ + $ay + 3 = 0$ and $y^2$ + $y + b = 0$ , then find the value of $ab$

A
$3$

B
$4$

C
$5$

D
$6$
Question 2
1 / -0

The ratio of the roots of the equation ${ ax }^{ 2 }+bx+c=0$ is same as the ratio of the roots of equation ${ px }^{ 2 }+qx+r=0$ . If ${ D }_{ 1 }$ and ${ D }_{ 2 }$ are the discriminants of ${ ax }^{ 2 }+bx+c=0$ and ${ px }^{ 2 }+qx+r=0$ respectively, then ${ D }_{ 1 }:{ D }_{ 2 }=$

A
$\dfrac { { b }^{ 2 } }{ { q }^{ 2 } }$

B
$\dfrac { b }{ q }$

C
$\dfrac { b }{ { q }^{ 2 } }$

D
none of these

Solution

$\textbf{Step 1: Using quadratic formula,}$

$\boldsymbol{x=\dfrac{-b\pm\sqrt{D}}{2a}}$

$ax^2+bx+c=0$

$x=\dfrac{-b\pm\sqrt{D_1}}{2a}$ $[\text{ Using Quadratic Formula }]$

$px^2+qx+r=0$

$x=\dfrac{-q\pm\sqrt{D_2}}{2a}$ $[\text{ Using Quadratic Formula }]$

$\text{As per given question, We have }$

$\dfrac{-b+\sqrt{D_1}}{-b-\sqrt{D_1}}=\dfrac{-q+\sqrt{D_2}}{-q-\sqrt{D_2}}$

$\textbf{Step 2: Using Componendo and Dividendo, we get}$

$\dfrac{-\sqrt{D_1}}{b}=\dfrac{-\sqrt{D_2}}{q}$

$\text{On squaring both side}$

$\dfrac{D_1}{b^2}=\dfrac{D_2}{q^2}$

$\Rightarrow \dfrac{D_1}{D_2}=\dfrac{b^2}{q^2}$
Question 3
1 / -0

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

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

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

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

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

Solution

Since roots of given quadratic are real, $D \geq 0$
$\Rightarrow (a+b+c)^2-3\lambda(ab+bc+ca) \geq 0$
$\Rightarrow \lambda \leq \cfrac{1}{3} \cfrac{(a+b+c)^2}{ab+bc+ca}=\cfrac{2}{3}+\cfrac{1}{3}\cfrac{a^2+b^2+c^2}{ab+bc+ca}$
Given $a,b,c$ are sides of triangle $\Rightarrow a+b>c\Rightarrow ac+bc>c^2$
Similarly, $ab+bc> b^2$ and $ca+ab> a^2$
Using these $a^2+b^2+c^2<2(ab+bc+ca)$
Hence, $\lambda < \cfrac{2}{3}+\cfrac{2}{3}=\cfrac{4}{3}$ .
Question 4
1 / -0

The equation $\sqrt {2x-2x^2-5}=x^2-2x-3$ , where $x$ is real has

A
No real solution

B
Exactly one real solution

C
Exactly two real solutions

D
Exactly four real solutions

Solution

Let us consider the expression, $2x-2x^2-5$ .
$\Delta = 2^2 - 4(-2)(-5)$
$\Delta = 4 - 40$
$\Delta = -36 < 0$
The coefficient of $x^2$ is negative. Hence the expression is negative for all $x$ .
Hence, the term inside the square root is always negative. So, there are no real roots for the equation.
Question 5
1 / -0

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

A
Can be the sides of a triangle

B
Cannot be the sides of triangle

C
Nothing can be said

D
None of these

Solution

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

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

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

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

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

$\sqrt{a}<\sqrt{b}+\sqrt{c}$
Hence sum of two sides is greater than the third side.
Hence a triangle can be formed with sides $\sqrt{a},\sqrt{b},\sqrt{c}$ .
Question 6
1 / -0

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

A
$4$

B
$5$

C
$4.5$

D
$7$

Solution

$\because x^2-6x+13=x^2-6x+9+4$
$=(x-3)^2+4\ge4$
because $(x-3)^2$ can never be less than zero
$\therefore$ least value of $x^2-6x+13=4$
Question 7
1 / -0

On solving $4x^2 - 4a^2x + (a^4 - b^4)$ = 0 we get value of $x$ is equal to

A
$\displaystyle\frac{a^2 \pm b^2}{2}$

B
$\displaystyle\frac{a^2}{2}$

C
$\displaystyle\frac{b^2}{2}$

D
$\displaystyle\frac{a^2 \div b^2}{2}$

Solution

On comparing the equation with $Ax^2+Bx+C = 0$ ,
We get, $A = 4, B = -4a^2, C = a^4-b^4$
D = $B^2 - 4AC$ = $(-4a^2)^2$ $-$ 4*4* $(a^4 - b^4)$
= $16a^4 - 16a^4 + 16b^4$ = $16b^4$

$x = \displaystyle\frac{- B \pm\sqrt D}{2A}$ = $\displaystyle\frac{- 4(-4a^2)\pm\sqrt {16b^4}}{2 \times 4}$

= $\displaystyle\frac {4a^2\pm4b^2}{8}$ = $\displaystyle\frac {a^2 + b^2}{2}$ or $\displaystyle\frac {a^2 - b^2}{2}$
Question 8
1 / -0

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

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

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

C
$0$

D
$1$

Solution

$\Rightarrow$ $\alpha$ and $\beta$ are the roots of quadratic equation $ax^2+2bx+c=0$ .
$\Rightarrow$ $\alpha\beta=\dfrac{c}{a}$ ----- ( 1 )
$\Rightarrow$ $\alpha+\beta=\dfrac{-2b}{a}$
$\Rightarrow$ $(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta$
$\Rightarrow$ $\left(\dfrac{-2b}{a}\right)^2=\alpha^2+\beta^2+2\times\dfrac{c}{a}$
$\therefore$ $\alpha^2+\beta^2=\dfrac{4b^2}{a^2}-\dfrac{2c}{a}$ ----- ( 2 )
$\Rightarrow$ $(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta$
$\Rightarrow$ $(\alpha-\beta)^2=\dfrac{4b^2}{a^2}-\dfrac{2c}{a}-\dfrac{2c}{a}$ [ From ( 1 ) and ( 2 ) ]
$\Rightarrow$ $(\alpha-\beta)^2=\dfrac{4b^2-4ac}{a^2}$
$\therefore$ $\alpha-\beta=\sqrt{\dfrac{4b^2-4ac}{a^2}}$ ------ ( 3 )
$\Rightarrow$ Now, $\alpha+\delta$ and $\beta+\delta$ are roots of quadratic equation $Ax^2+2Bx+C=0$
$\Rightarrow$ $(\alpha+\delta)(\beta+\delta)=\dfrac{C}{A}$
$\Rightarrow$ $\alpha\beta+\alpha\delta+\beta\delta+\delta^2=\dfrac{C}{A}$
$\therefore$ $\alpha\beta=\dfrac{C}{A}-\alpha\delta-\beta\delta-\delta^2$ -------- ( 4 )
$\Rightarrow$ $(\alpha+\delta)+(\beta+\delta)=\dfrac{-2B}{A}$
$\Rightarrow$ $\alpha+\beta+2\delta=\dfrac{-2B}{A}$
$\Rightarrow$ $\alpha+\beta=\dfrac{-2B}{A}-2\delta$ -------- ( 5 )
$\Rightarrow$ $(\alpha+\beta)^2=\left(\dfrac{-2B}{A}-2\delta\right)^2$
$\therefore$ $\alpha^2+\beta^2=\left(\dfrac{-2B}{A}-2\delta\right)^2-2\alpha\beta$ -------- ( 6 )
$\Rightarrow$ $[(\alpha+\delta)-(\beta+\delta)]^2=(\alpha-\beta)^2$
$\Rightarrow$ $(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta$
$\Rightarrow$ $(\alpha-\beta)^2=\left(\dfrac{-2B}{A}-2\delta\right)^2-2\alpha\beta-2\alpha\beta$ [ From ( 6 ) ]
$\Rightarrow$ $(\alpha-\beta)^2=\left(\dfrac{-2B}{A}-2\delta\right)^2-4\alpha\beta$
$\Rightarrow$ $(\alpha-\beta)^2=\dfrac{4B^2}{A^2}+\dfrac{8B\delta}{A}+4\delta^2-\dfrac{4C}{A}+4\alpha\delta+4\beta\delta+4\delta^2$ [ From ( 4 ) ]
$\Rightarrow$ $(\alpha-\beta)^2=\dfrac{4B}{A^2}-\dfrac{4C}{A}+\dfrac{8B\delta}{A}+8\delta^2+4\delta(\alpha+\beta)$
$\Rightarrow$ $(\alpha-\beta)^2=\dfrac{4B^2-4AC}{A^2}+\dfrac{8B\delta}{A}+8\delta^2-\dfrac{8B\delta}{A}-8\delta^2$ [ From ( 5 ) ]
$\Rightarrow$ $(\alpha-\beta)^2=\dfrac{4B^2-4AC}{A^2}$

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

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

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

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

$\therefore$ $\dfrac{b^2-ac}{B^2-AC}=\dfrac{a^2}{A^2}=\left(\dfrac{a}{A}\right)^2$
Question 9
1 / -0

The roots of the equation $\displaystyle \left ( x-a \right )\left ( x-b \right )+\left ( x-b \right )\left ( x-c \right )+\left ( x-c \right )\left ( x-a \right )=0$ are

A
positive

B
negative

C
real

D
imaginary

Solution

Given equation is written as
$\displaystyle 3x^{2}-2x\left ( a+b+c \right )+\left ( ab+bc+ca \right )=0$
Discriminant is $\displaystyle 4\left ( a+b+c \right )^{2}-12\left ( ab+bc+ca \right )$
$\displaystyle =4\left ( a^{2}+b^{2}+c^{2}-ab-bc-ca\right )$
$\displaystyle =2\left [ \left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2} \right ]$
=+ve
$\displaystyle \therefore$ Roots are real
Question 10
1 / -0

If the quadratic equation $\displaystyle x^{2}-mx-4x+1=0$ has real and distinct roots, then the values of $m$ are
A. $\displaystyle (-\infty , -6)$
B. $\displaystyle (-\infty,-3)$
C. $(-2,\infty )$
D. $\displaystyle \left ( 2,\infty \right )$

A
A or C

B
A or D

C
B or C

D
B or D

Solution

For an equation $a{x}^{2} + bx + c = 0$ , the discriminant $\triangle = {b}^{2} -4ac$ helps us understand the nature of the roots.

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

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

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

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

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

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

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

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

$\Rightarrow$ $m > -6 ; m > -2$

From these two solutions, we have $m > -2$

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

$\Rightarrow$ $m < -6 ; m< -2$

From these two solutions, we have $m <- 6$

So both $A$ and $C$ are the possible values of $m$

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Quadratic Equations Test - 43

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Quadratic Equations Test - 43

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Question 1

333

444

555

666

Question 2

b2q2 \dfrac { { b }^{ 2 } }{ { q }^{ 2 } } q2b2​

bq \dfrac { b }{ q } qb​

bq2 \dfrac { b }{ { q }^{ 2 } } q2b​

none of these

Solution

Question 3

λ<43\displaystyle \lambda <\dfrac{4}{3}λ<34​

λ>53\displaystyle \lambda >\frac{5}{3}λ>35​

λ∈(13,53)\displaystyle \lambda \in \left ( \frac{1}{3}, \frac{5}{3} \right )λ∈(31​,35​)

λ∈(43,53)\displaystyle \lambda \in \left ( \frac{4}{3}, \frac{5}{3} \right )λ∈(34​,35​)

Solution

Question 4

No real solution

Exactly one real solution

Exactly two real solutions

Exactly four real solutions

Solution

Question 5

Can be the sides of a triangle

Cannot be the sides of triangle

Nothing can be said

None of these

Solution

Question 6

444

555

4.54.54.5

777

Solution

Question 7

a2±b22\displaystyle\frac{a^2 \pm b^2}{2}2a2±b2​

a22\displaystyle\frac{a^2}{2}2a2​

b22\displaystyle\frac{b^2}{2}2b2​

a2÷b22\displaystyle\frac{a^2 \div b^2}{2}2a2÷b2​

Solution

Question 8

(aA)2\displaystyle \left ( \frac{a}{A} \right )^{2}(Aa​)2

(Aa)2\displaystyle \left ( \frac{A}{a} \right )^{2}(aA​)2

000

111

Solution

Question 9

positive

negative

real

imaginary

Solution

Question 10

A or C

A or D

B or C

B or D

Solution

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$3$

$4$

$5$

$6$

$\dfrac { { b }^{ 2 } }{ { q }^{ 2 } }$

$\dfrac { b }{ q }$

$\dfrac { b }{ { q }^{ 2 } }$

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

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

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

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

$4$

$5$

$4.5$

$7$

$\displaystyle\frac{a^2 \pm b^2}{2}$

$\displaystyle\frac{a^2}{2}$

$\displaystyle\frac{b^2}{2}$

$\displaystyle\frac{a^2 \div b^2}{2}$

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

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

$0$

$1$