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

$$3$$

$$4$$

$$5$$

$$6$$

Question 2

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

$$ \dfrac { b }{ q } $$

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

none of these

Solution

Question 3

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

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

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

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

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

$$4$$

$$5$$

$$4.5$$

$$7$$

Solution

Question 7

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

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

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

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

Solution

Question 8

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

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

$$0$$

$$1$$

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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