Question

Let $a,b\in R$, $a\ne 0$, such that the equation, $a{x}^2-2bx+5=0$ has a repeated root $\alpha$, which is also a root of the equation$x^2-2bx-10=0$. If $\beta$ is the root of this equation, then ${\alpha }^2+{\beta }^2$ is equal to:

• A. $24$
• B. $25$
• C. $26$
• D. $28$

Answer 1

The correct option is B

$25$

Explanation for correct option:

Step:1:Finding the value of $\alpha$

If $\alpha$and $\beta$be the root of the equation $a{x}^2+bx+c=0$, then

Sum of roots, $\alpha +\beta =-\frac{b}{a}$ and

Product of roots, $\alpha \beta =\frac{c}{a}$.

$a{x}^2-2bx+5=0$ has a repeated root $\alpha$, i.e both roots of this equation is $\alpha$.

Equating the sum of roots,

$$\begin{gathered} \Rightarrow \alpha +\alpha =-\left(\frac{-2b}{a}\right) \\ \Rightarrow 2\alpha =\frac{2b}{a} \\ \Rightarrow \alpha =\frac{b}{a}\dots \left(i\right) \end{gathered}$$

Also, by product of roots

$$\begin{gathered} \Rightarrow \alpha \times \alpha =\frac{5}{a} \\ \Rightarrow {\alpha }^2=\frac{5}{a}\dots \left(ii\right) \end{gathered}$$.

Substituting equation $\left(i\right)\text{in}\left(ii\right)$, to get a relation between $b$ and $a$.

$$\begin{gathered} \Rightarrow {\left(\frac{b}{a}\right)}^2=\frac{5}{a}\left[\because \alpha =\frac{b}{a}\right] \\ \Rightarrow b^2=5a\left[\because a\ne 0\right]\to \left(iii\right) \end{gathered}$$

The given equation$x^2-2bx-10=0$ is, then $\alpha +\beta =2b$ , $\alpha \beta =-10$

Given, $\alpha =\frac{b}{a}$be one of the roots of this equation. Substituting this in $a{x}^2-2bx+5=0$

$$\begin{gathered} \Rightarrow {\left(\frac{b}{a}\right)}^2-2b\left(\frac{b}{a}\right)-10=0 \\ \Rightarrow \frac{b^2}{a^2}-\frac{2b^2}{a}-10=0 \\ \Rightarrow b^2-2b^{2}a-10a^2=0 \\ \Rightarrow 5a-2\left(5a\right)a-10a^2=0\left[\text{by equation}\left(\mathrm{iii}\right)\because b^2=5a\right] \\ \Rightarrow 5a-10a^2-10a^2=0 \\ \Rightarrow 5a=20a^2 \\ \Rightarrow a=\frac{1}{4} \end{gathered}$$

$\Rightarrow b^2=5a=\frac{5}{4}(\text{from}(iii\left)\right)$

Step:2 Finding the value of $\beta$ and ${\alpha }^2+{\beta }^2$

$$\begin{gathered} \Rightarrow {\alpha }^2=\frac{5}{a}=\frac{5}{\left({\displaystyle \frac{1}{4}}\right)}=20 \\ \Rightarrow \alpha \beta =-10\mathbf{}\left(\text{from the equation}\right) \\ \Rightarrow {\left(\alpha \beta \right)}^2={\left(-10\right)}^2\mathbf{}\mathbf{\text{(squaring both sides}}\text{)} \\ \Rightarrow \left(20\times {\beta }^2\right)=100 \\ \Rightarrow {\beta }^2=\frac{100}{20}=5 \end{gathered}$$

Hence, the value of ${\alpha }^2+{\beta }^2=20+5=25$

Thus, option (B) is the correct answer.