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Let $\mathrm{a,b\in \mathbb{R}}$ be such that the equation $\mathrm{ax^{2}-2bx+15=0}$ has a repeated root $\mathrm{a}$ . If $\mathrm{a}$ and $\mathrm{\beta }$ are the roots of the equation $\mathrm{x^{2}-2bx+21=0}$ , then $\mathrm{a^{2}+\beta ^{2}}$ is equal to:

Option: 1
$37$

Option: 2
$58$

Option: 3
$68$

Option: 4
$92$

Answers (1)

best_answer

Given

$\mathrm{ax^{2}-2bx+15=0}$

Has repeated roots so $\mathrm{D=0}$

$\mathrm{4 b^{2}-4 \times 15 \times a=0}\\$

$\mathrm{\Rightarrow b^{2}=15 a}$

Also Given

$\mathrm{x^{2}-2 b x+21=0}$

now $\mathrm{\alpha }$ will satisfy both qudratic

$\mathrm{a x^{2}-2 b x+15=0 \text { and } x^{2}-2 b x+21=0 }$

Putting the value we get

$\mathrm{a \alpha^{2}-2 b \alpha+15=0 }\\$

$\mathrm{-\alpha^{2}-2 b \alpha+21=0}\\$

_______________________

$\mathrm{(a-1) \alpha^{2}=6}$

$\mathrm{\alpha^{2}=\frac{6}{a-1}}\\$

Now in eqn (1) product of root $\mathrm{ \alpha^{2}=\frac{15}{a}}\\$

$\mathrm{ \text { so } \frac{15}{a}=\frac{6}{a-1} \Rightarrow 2 a=5 a-5 \Rightarrow a=\frac{5}{3}}$

$\mathrm{\text { Now } b^{2}=15 a \Rightarrow b^{2}=15 \times \frac{5}{3} \Rightarrow b^{2}=25}$

$\mathrm{ \text { so } b=\pm 5}$

Now in quadratic $\mathrm{ x^{2}-2bx+21=0}$

Putting the value of b we get

$\mathrm{x^{2}-10 x+21=0 \Rightarrow(x-7)(x-3) =0}\\$

$\mathrm{so\: x=3 \: or\: x=7}\\$

$\mathrm{\text { or } x^{2}+10 x+21=0 \Rightarrow x=-3 \text { or } x=-7 }\\$

$\mathrm{\text { so } \alpha=\pm 3 \text { \& } \beta=\pm 7}\\$

$\mathrm{\text { so } \alpha^{2}+\beta^{2}=3^{2}+7^{2}=9+49=58}$

Hence the correct answer is option 2

Posted by

Ritika Harsh

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