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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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