QUADRATIC EQUATION

 Question 1:

If $\alpha \ne \beta$ but ${\alpha }^2=5\alpha -3$ and ${\beta }^2=5\beta -3$ then the equation whose roots are $\alpha /\beta$ and $\beta /\alpha$ is

[1] $3x^2-25x+3=0$

[2] $x^2+5x-3=0$

[3] $x^2-5x+3=0$

[4] $3x^2-19x+3=0$

Answer & Solution

Option # 4

We need the equation whose roots are $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ which are reciprocal of each other, which means product of roots is $\frac{\alpha }{\beta }\frac{\beta }{\alpha }=1.$ In our choice (a) and (d) have product of roots 1, so choices (b) and (d) are out of court. In the problem choice, None of these is not given. If out of four choices only one  of choice satisfies that product of root is 1 then you select that choice for correct answer.

Now for proper choice we proceed as,$\alpha \ne \beta ,$ but ${\alpha }^2=5\alpha -3$ and ${\beta }^2=5\beta -3$

changing $\alpha ,\beta$ by $x$

Therefore, $\alpha ,\beta$ are roots of $x^2-5x+3=0$

$\Rightarrow \alpha +\beta =5,\alpha \beta =3$

now, $S=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{19}{3}$ and product

$\frac{\alpha }{\beta }\cdot \frac{\beta }{\alpha }=1$

Therefore Required equation,

$x^2-($ sum of roots $)x+$ product of roots $=0$

$\Rightarrow x^2-\frac{19}{3}x+1=0$

$\Rightarrow 3x^2-19x+3=0$ is correct answer.

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Question 2:
Difference between the corresponding roots of $x^2+ax+b=0$ and $x^2+bx+a=0$ is same and $a\ne b,$ then

[1] $a+b+4=0$

[2] $a+b-4=0$

[3] $a-b-4=0$

[4] $a-b+4=0$

Answer & Solution

Option # 1

Let $\alpha ,\beta$ are roots of $x^2+bx+a=0$

Therefore $\alpha +\beta =-b$ and $\alpha \beta =a$

again let $\gamma ,\delta$ are roots of $x^2+ax+b=0$

Therefore $\gamma +\delta =-a$ and $\gamma \delta =b$

Now given

$\alpha -\beta =\gamma -\delta$

$\Rightarrow (\alpha -\beta {)}^2=(\gamma -\delta {)}^2$

$\Rightarrow (\alpha +\beta {)}^2-4\alpha \beta =(\gamma +\delta {)}^2-4\gamma \delta$

$\Rightarrow b^2-4a=a^2-4b$

$\Rightarrow b^2-a^2=-4(b-a)$

$\Rightarrow (b-a)(b+a+4)=0$

$\Rightarrow b+a+4=0$ as $(a\ne b)$

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Question 3:
If p and q are the roots of the equation $x^2+px+q=0,$ then

[1] $p=1,q=-2$

[2] $p=0,q=1$

[3] $p=-2,q=0$

[4] $p=-2,q=1$

Answer & Solution

Option # 1

Given $S=p+q=-p$ and product $pq=q$

$\Rightarrow q(p-1)=0$

$\Rightarrow q=0,p=1$

Now If $q=0$ then $p=0\Rightarrow p=q$

If $p=1,$ then $p+q=-p$

$q=-2p$

$q=-2\left(1\right)$

$q=-2$

$\Rightarrow p=1$ and $q=-2$

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Question 4:
If a , b , c are distinct positive real numbers and $a^2+b^2+c^2=1$ then $ab+bc+ca$ is

[1] less than 1

[2] equal to 1

[3] greater than 1

[4] any real no

Answer & Solution

Option # 1

In such type of problem if sum of the squares of number is known and we needed product of numbers taken two at a time or needed range of the product of numbers taken two at a time. We start square of the sum of the numbers like

$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

$\Rightarrow 2(ab+bc+ca)=(a+b+c{)}^2-(a^2+b^2+c^2)$

$\Rightarrow ab+bc+ca=\frac{(a+b+c{)}^2-1}{2}<1$

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Question 5:
The value of a for which one root of the quadratic equation $(a^2-5a+3)x^2+(3a-1)x+2=0$ is twice as large as the other is

[1] -2/3

[2] 1/3

[3] -1/3

[4] 2/3

Answer & Solution

Option # 4

Let $\alpha ,2\alpha$ are roots of the given equation therefore sum of the roots

$\alpha +2\alpha =3\alpha =\frac{1-3a}{a^2-5a+3}\dots \left(1\right)$

and product of roots

$\alpha \left(2\alpha \right)=2{\alpha }^2=\frac{2}{a^2-5a+3}\cdots \left(2\right)$

By (i) and (ii) we have

$\frac{9{\alpha }^2}{2{\alpha }^2}=\frac{(1-3a{)}^2}{{(a^2-5a+3)}^2}\times \frac{a^2-5a+3}{2}$

$\Rightarrow 9(a^2-5a+3)=(1-3a{)}^2$

$\Rightarrow a=\frac{2}{3}$

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Question 6:
If the sum of the roots of the quadratic equation $a{x}^2+bx+c=0$ is equal to the sum of the squares of their reciprocals, then $\frac{a}{c},\frac{b}{a}$ and $\frac{c}{b}$ are in

[1] geometric progression

[2] harmonic progression

[3] arithmetic-geometric progression

[4] arithmetic progression

Answer & Solution

Option # 2

Given $\alpha +\beta$

$=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{{\alpha }^2{\beta }^2}$

$\Rightarrow 2a^{2}c=b{c}^2+a{b}^2$

$\Rightarrow \frac{2a}{b}=\frac{c}{a}+\frac{b}{c}$

$\Rightarrow \frac{c}{a},\frac{a}{b},\frac{b}{c}\in A.P$

$\Rightarrow$ reciprocals are in H.P

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Question 7:
The number of real solutions of the equation $x^2-3|x|+2=0$ is

[1] 4

[2] 1

[3] 3

[4] 2

Answer & Solution

Option # 1

Given $x^2-3|x|+2=0$

If $x\ge 0$ i.e. $|x|=x$

Therefore,  the given equation can be written as

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

$\Rightarrow (x-1)(x-2)=0$

$\Rightarrow x=1,2$

Similarly for $x<0,x^2-3|x|+2=0$

$\Rightarrow x^2+3x+2=0$

$\Rightarrow x=-1,-2$

Hence $1,-1,2,-2$ are four solutions of the given equation.

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Question 8:
Let two numbers have arithmetic mean 9 and geometric mean 4 . Then these numbers are the roots of the quadratic equation

[1] $x^2+18x-16=0$

[2] $x^2-18x+16=0$

[3] $x^2+18x+16=0$

[4] $x^2-18x-16=0$

Answer & Solution

Option # 2

Let the two number be $\alpha ,\beta$

Therefore, $\frac{\alpha +\beta }{2}=9$ and $\sqrt{\alpha \beta }=4$

Therefore,  Required equation

$x^2-2($ Average value of $\alpha ,\beta )x+{\sqrt{GM}}^2=0$

$x^2-2\left(9\right)x+16=0$

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Question 9:
If $(1-p)$ is a root of quadratic equation $x^2+px+(1-p)=0$ then its roots are

[1] 0, -1

[2] -1, 1

[3] 0, 1

[4] -1, 2

Answer & Solution

Option # 1

$1-p$ is root of $x^2+px+1-p=0$

$\Rightarrow (1-p{)}^2+p(1-p)+(1-p)=0$

$(1-p)[1-p+p+1]=0$

$\Rightarrow p=1$

Given equation becomes $x^2+x=0$

$\Rightarrow x=0,-1$

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Question 10:
If one root of the equation $x^2+px+12=0$ is 4 while the equation $x^2+px+q=0$ has equal roots, then the value of q is

[1] 3

[2] 12

[3] 49/4

[4] 4

Answer & Solution

Option # 3

 As $x^2+px+q=0$ has equal roots $:p^2=4q$

and one root of $x^2+px+12=0$ is 4

$16+4p+12=0$

OR $p=-7$

$p^2=4q\Rightarrow q=\frac{49}{4}$

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Question 11:
If the roots of the equation $x^2-bx+c=0$ be two consecutive integers, then $b^2-4c$ equals

[1] 3

[2] -2

[3] 1

[4] 2

Answer & Solution

Option # 1

Let $\alpha ,\alpha +1$ are consecutive integer

Therefore, $(x+\alpha )(x+\alpha +1)=x^2-bx+c$ comparing both sides we get $\Rightarrow -b=2\alpha +1$

$c={\alpha }^2+\alpha$

$b^2-4c=(2\alpha +1{)}^2-4({\alpha }^2+\alpha )=1$

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Question 12:
If both the roots of the quadratic equation $x^2-2kx+k^2+k-5=0$ are less than 5 then k lies in the interval

[1] $(6,\infty )$

[2] $(5,6]$

[3] [ 4,5 ]

[4] $(-\infty ,4)$

Answer & Solution

Option # 4

Given $x^2-2kx+k^2+k-5=0$

Roots are less than $5\Rightarrow D\ge 0$

$\Rightarrow (-2k{)}^2\ge 4(k^2+k-5)\Rightarrow k\le 5\cdots (A)$

Again $f\left(5\right)>0$

$\Rightarrow 25-10k+k^2+k-5>0$

$\Rightarrow k^2-9k+20>0\Rightarrow (k-4)(k-5)>0$

$\Rightarrow k<4\cup k>5\cdots \left(B\right)$

Also $\frac{\text{sum of roots}}{2}<5\Rightarrow k<5\cdots \left(C\right)$

from $\left(\text{A}\right),\left(\text{B}\right),\left(\text{C}\right)$ we have

$k\in (-\infty ,4)$ as the choice gives number $k<5$

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Question 13:
If the equation $a_n{x}^n+a_{n-1}{x}^{n-1}+\dots +a_{1}x=0$ $a_1\ne 0,n\ge 2,$ has a positive root $x=\alpha$ , then the equation $n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+\dots +a_1=0$ has a positive root, which is

[1] smaller than $\alpha$

[2] greater than $\alpha$

[3] equal to $\alpha$

[4] greater than or equal to $\alpha$

Answer & Solution

Option # 1

If possible say

$f\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+\dots +a_{n}x$

therefore f ( 0 ) = 0

Now $f\left(\alpha \right)=0$ ( therefore x = $\alpha$ is root of given equation )

Therefore, $f^{\prime }\left(x\right)=n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+\dots +a_1=0$

has at least one root in $]0,\alpha [$

$\Rightarrow n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+\dots +a_1=0$

has a +root smaller than $\alpha .$

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Question 14:
All the values of $m$ for which both roots of the equation $x^2-2mx+m^2-1=0$ are greater than - 2 but less than 4 , lie in the interval

[1] $-2<m<0$

[2] $m>3$

[3] $-1<m<3$

[4] $1<m<4$

Answer & Solution

Option # 3

Let $\alpha ,\beta$ are roots of the equation $(x^2-2mx+m^2)=1$

$\Rightarrow x=m\pm 1=m+1,m-1$

Now $-2<m+1<4\cdots \left(1\right)$

and $-2<m-1<4\cdots \left(2\right)$

$\{\begin{array}{cc}\Rightarrow & -3<m<3\cdots \left(A\right)\\ \text{and}& -1<m<5\cdots \left(B\right)\end{array}$

By (A) and (B) we get $-1<m<3$ as shown by the number line.

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Question 15:
If the difference between the roots of the equation $x^2+ax+1=0$ is less than $\sqrt{5}$,  then the set of possible values of a is

[1] $(3,\infty )$

[2] $(-\infty ,-3)$

[3] $(-3,3)$

[4] $(-3,\infty )$

Answer & Solution

Option # 3

$x^2+ax+1=0$

Let roots be $\alpha$ and $\beta ,$ then $\alpha +\beta =-a$ and $\alpha \beta =1$

$|\alpha -\beta |=\sqrt{(\alpha +\beta {)}^2-4\alpha \beta },|\alpha -\beta |=\sqrt{a^2-4}$

since, $|\alpha -\beta |<\sqrt{5}\Rightarrow \sqrt{a^2-4}<\sqrt{5}$

$\Rightarrow a^2-4<5\Rightarrow a^2<9\Rightarrow -3<a<3$

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Question 16:
The quadratic equations $x^2-6x+a=0$ and $x^2-cx+6=0$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3 . Then the common root is

[1] 2

[2] 1

[3] 4

[4] 3

Answer & Solution

Option # 1

Let $\alpha$ and 4$\beta$ be the root of

$x^2-6x+a=0$

and $\alpha$ and 3$\beta$ be those of the equation

$x^2-cx+6=0$

From the relation between roots and coefficients

$\alpha +4\beta =6$ and $4\alpha \beta =a$

$\alpha +3\beta =c$ and $3\alpha \beta =6$

we obtain $\alpha \beta =2$ giving $a=8$

The first equation is $x^2-6x+8=0\Rightarrow x=2,4$

For $\alpha =4,4\beta =2\Rightarrow 3\beta =3/2$ (not an integer)

So the common root is $\alpha =2$