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Let $a,b\epsilon \textbf{R},a\neq 0$ be such that the equation, $ax^{2}-2bx+5=0$ has a repeated root $\alpha$ , which is also a root of the equation, $x^{2}-2bx-10=0$ . If $\beta$ is the other root of this equation, then $\alpha ^{2}+\beta ^{2}$ is equal to:
Option: 1 $24$
Option: 2 $25$
Option: 3 $26$
Option: 4 $28$

Answers (1)

best_answer

Nature of Roots -

Let the quadratic equation is ax² + bx + c = 0

D is the discriminant of the equation.

iii) if roots D = 0, then roots will be real and equal, then

$\\\mathrm{x_1=x_2 = \frac{-b}{2a} } \\\mathrm{Then, \;\; ax^2+bx +c =a(x-x_1)^2 }$

-

ax² – 2bx + 5 = 0 having equal roots or $D=0$ and $\alpha=\frac{b}{a}$

$(2b)^2=4\times5\times a\;\;\Rightarrow \;\;b^2=5a$

Put $\alpha=\frac{b}{a}$ in the second equation

${x^{2}-2 b x-10=0} \\ {\Rightarrow b^{2}-2 a b^{2}-10 a^{2}=0}$

$\\\Rightarrow 5 a-10 a^{2}-10 a^{2}=0 \\ \Rightarrow 20 a^{2}=5 a \\ \Rightarrow a=\frac{1}{4} \text { and } \mathrm{b}^{2}=\frac{5}{4} \\ \alpha^{2}= 20 \text { and } \beta^{2}=5 \\ \alpha^{2}+\beta^{2} \\ =5+20 \\ =25$

Correct Option 2

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avinash.dongre

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