Quadratic Equations Questions and answers

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Elementary Mathematics

31). If \( \Large \alpha , \beta \) are the roots of equation \( \Large ax^{2}+bx+c=0 \), then \( \Large \frac{ \alpha }{ \alpha \beta +b}+\frac{ \beta }{ \alpha x+b} \) is equal to

A). 2/a

B). 2/b

C). 2/c

D). -2/a

View Answer

Correct Answer: -2/a

Since, \( \Large \alpha and\ \beta \) are the roots of the equation \( \Large ax^{2}+bx+c=0 \)

\( \Large \alpha + \beta = -\frac{b}{a}, \alpha \beta = \frac{c}{a} \)

and \( \Large \alpha ^{2}+ \beta ^{2}= \left( \alpha + \beta \right)^{2}-2 \alpha \beta = \left(\frac{b^{2}-2ac}{a^{2}}\right) \)

\( \Large \frac{ \alpha }{ \alpha \beta +b}+\frac{ \beta }{a \alpha +b}= \frac{a \left(a \alpha +b\right)+ \beta \left(a \beta +b\right) }{ \left(a \beta +b\right) \left(a \alpha +b\right) } \)

= \( \Large \frac{a \left( \alpha ^{2}+ \beta ^{2}\right)+b \left( \alpha + \beta \right) }{ \alpha \beta a^{2}+ab \left( \alpha + \beta \right)+b^{2} } \)

= \( \Large \frac{a \left(\frac{b^{2}-2ac}{a^{2}}\right)+b \left(-\frac{b}{a}\right) }{ \left(\frac{c}{a}\right)a^{2}+ab \left(-\frac{b}{a}\right)+b^{2} } \)

= \( \Large \frac{b^{2}-2ac-b^{2}}{a^{2}c-ab^{2}+ab^{2}}=\frac{-2ac}{a2c}=\frac{-2}{a} \)

32). If the roots of the equation \( \Large \frac{ \alpha }{x- \alpha }+\frac{ \beta }{x- \beta }=1 \) be equal in magnitude but opposite in sign, then \( \Large \alpha + \beta \) is equal to:

A). 0

B). 1

C). 2

D). none of these

View Answer

Correct Answer: 0

Given equation \( \Large \frac{ \alpha }{x- \alpha }+\frac{ \beta }{x- \beta }=1 \) can be rewritten as \( \Large x^{2}-2 \left( \alpha + \beta \right)x+3 \alpha \beta =0 \)

Let roots be \( \Large \alpha' and\ - \alpha' \)

=> \( \Large \alpha' + \left(- \alpha' \right)=2 \left( \alpha + \beta \right) \)

=> \( \Large 0=2 \left( \alpha + \beta \right) \)

=> \( \Large \alpha + \beta = 0 \)

33). If the roots of the equation \( \Large px^{2}+2qx+r=0 \) and \( \Large qx^{2}-2\sqrt{pr}x+q=0 \) be real, then:

A). \( \Large p=q \)

B). \( \Large q^{2}=pr \)

C). \( \Large p^{2}=qr \)

D). \( \Large r^{2}=pq \)

View Answer

Correct Answer: \( \Large q^{2}=pr \)

Given equations are \( \Large px^{2}+2qx+r=0 \) and \( \Large qx^{2}-2\sqrt{pr}x+q=0 \)

Since, they have real roots

\( \Large 4q^{2}-4pr\ge 0 => q^{2}\ge pr..\) ...(i)

and from second, \( \Large 4 \left(pr\right)-4q^{2}\ge 0 \)

=> \( \Large pr\ge q^{2} \)

From equations (i) and (ii) we get

\( \Large q^{2}=pr \)

34). If one root of the quadratic equation \( \Large ax^{2}+bx+c=0 \) is equal to nth power of the other root, then the value of \( \Large \left(ac^{n}\right)^{\frac{1}{n+1}} + \left(a^{n}c\right)^{\frac{1}{n+1}} \) is equal to

A). b

B). -b

C). \( \Large b^{1/n+1} \)

D). \( \Large -b^{1/n+1} \)

View Answer

Correct Answer: -b

Let \( \Large a, a^{n} \) be the roots of equation \( \Large ax^{2}+bx+c=0 \)

then \( \Large a+a^{n}=\frac{-b}{a} and\ a.a^{n}=\frac{c}{a} => a^{n+1}=\frac{c}{a} \)

Eliminating \( \Large \alpha \), we get

\( \Large \left(\frac{c}{a}\right)^{\frac{1}{n+1}}+ \left(\frac{c}{a}\right)^{\frac{n}{n+1}} =-\frac{b}{a} \)

=> \( \Large a.a^{-\frac{1}{n+1}\frac{1}{c^{n+1}}}+a.a^{-\frac{n}{n+1}\frac{n}{n+1}}=-b \)

=> \( \Large \left(a^{n}c\right)^{\frac{1}{n+1}}+ \left(ac^{n}\right)^{\frac{1}{n+1}}=-b \)

35). The quadratic in t, such that AM of its roots is A and GM is G is:

A). \( \Large t^{2}-2At+G^{2}=0 \)

B). \( \Large t^{2}-2At-G^{2}=0 \)

C). \( \Large t^{2}+2At+G^{2}=0 \)

D). none of these.

View Answer

Correct Answer: \( \Large t^{2}-2At+G^{2}=0 \)

Let \( \Large \alpha \), \( \Large \beta \) are the root of the required equation,

then \( \Large A = \frac{ \alpha + \beta }{2} => \alpha \beta = 2A \)

and \( \Large G=\sqrt{ \alpha \beta } => \alpha \beta = G^{2} \)

the required equation is \( \Large t^{2}-2At+G^{2}=0 \)

36). If \( \Large n^{2}px+1 \) is a factor of expression \( \Large ax^{3}+bx+c \) then:

A). \( \Large a^{2}+c^{2}=-ab \)

B). \( \Large a^{2}-c^{2}=-ab \)

C). \( \Large a^{2}-c^{2}=ab \)

D). none of these

View Answer

Correct Answer: \( \Large a^{2}+c^{2}=-ab \)

Given that \( \Large x^{2}+px+1 \) is factor of \( \Large ax^{3}+bx+c=0 \) then

let \( \Large ax^{3}+bx+c= \left(x^{2}+px+1\right) \left(ax+\lambda\right) \), where \( \Large \lambda \) is constant. Then equating the coefficient of like powers of x on both sides, we get.

\( \Large 0 = ap+\lambda, b=p\lambda+a, c=\lambda \)

\( \Large p=\frac{-\lambda}{a}=\frac{-c}{a} \)

Hence, \( \Large b= \left(-\frac{c}{a}\right)c+a \) or \( \Large ab=a^{2}-c^{2} \)

37). If \( \Large \sqrt{3x^{2}-7x-30} + \sqrt{2x^{2}-7x-5} = x+5 \), then x is equal to:

A). 2

B). 3

C). 6

D). 5

View Answer

Correct Answer: 6

We have

\( \Large \sqrt{3x^{2}-7x-30}+\sqrt{2x^{2}-7x-5}=x+5 \)

or \( \Large \sqrt{3x^{2}-7x-30}= \left(x+5\right)-\sqrt{2x^{2}-7x-5} \)

on squaring both sides, we get

\( \Large 3x^{2}-7x-30=x^{2}+25+10x+ \left(2x^{2}-7x-5\right) \)

\( \Large -2 \left(x+5\right)\sqrt{2x^{2}-7x-5} => \sqrt{2x^{2}-7x-5}=5 \)

Again squaring

=> \( \Large 2x^{2}-7x-30=0 => x=6 \)

38). The value \( \Large 2+\frac{1}{2+\frac{1}{2+........\infty}} \)

A). \( \Large 1-\sqrt{2} \)

B). \( \Large 1+\sqrt{2} \)

C). \( \Large 1\pm \sqrt{2} \)

D). none of these

View Answer

Correct Answer: \( \Large 1+\sqrt{2} \)

Let \( \Large x=2+\frac{1}{2+\frac{1}{2+...\infty}} \)

=> \( \Large x = 2+\frac{1}{x} => x^{2}-2x-1=0 \)

=> \( \Large x=\frac{2\pm \sqrt{4+4}}{2} \)

=> \( \Large x=1\pm \sqrt{2} \)

But the value of given expression cannot be negative or less than 2, therefore \( \Large 1+\sqrt{2} \) is required answer.

39). If the roots of the equation \( \Large qx^{2}+px+q=0 \) are complex, where p, q are real then the roots of the equation \( \Large x^{2}-4qx+p^{2}=0 \) are:

A). real and unequal

B). real and equal

C). imaginary

D). none of these

View Answer

Correct Answer: real and unequal

The given equations are

\( \Large qx^{2}+px+q=0 \)

and \( \Large x^{2}-4qx+p_{2}=0 \)

Since, root of the equation (i) are complex, therefore \( \Large P^{2}-4q^{2}<0 \). Now, discriminant or equation (ii) is \( \Large 16q^{2}-4 P^{2}=-4 \left(P^{2}-4q^{2}\right)>0 \) Hence, roots are real and unequal.

40). The number or real solution of the equation \( \Large x^{2}-3|x|+2=0 \) is:

A). 2

B). 4

C). 1

D). 3

View Answer

Correct Answer: 4

Given equations is \( \Large x^{2}-3|x|+2=0 \). If x > 0 then lxl = x

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

=> \( \Large x^{2}-2x-x+2=0 \)

=> \( \Large x \left(x-2\right)-1 \left(x-2\right)=0 \)

=> \( \Large \left(x-1\right) \left(x-2\right)=0 \)

x = 1,2

If x < 0, then \( \Large |x|=-x \)

=> \( \Large x^{2}+3x+2=0 \)

=> \( \Large x^{2}+2x+x+2=0 \)

=> \( \Large x \left(x+2\right)+1 \left(x+2\right)=0 \)

=> \( \Large \left(x+1\right) \left(x+2\right)=0 \)

=> x = -1, -2

Hence four solutions are possible.

1	2	3	4	5	6	7

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