Q53 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
The equations $2 x^2+a x-2=0$ and $x^2+x+2 a=0$ have exactly one common root. If $a \neq 0$, then one of the roots of the equation $a x^2-4 x-2 a=0$ is
- A2
- B-2
- C$\frac{-4+\sqrt{22}}{3}$
- Correct answerD$\frac{-2+\sqrt{22}}{3}$
Solution
We know that condition of exactly one common root of quadratic equations $a_1 x^2+b_1 x+c_1=0$ and $a_2 x^2+b_2 x+c_2=0$ is
$\left(c_1 a_2-c_2 a_1\right)^2=\left(b_1 c_2-b_2 c_1\right)\left(a_1 b_2-a_2 b_1\right)$
Here, $a_1=2, b_1=a, c_1=-2, a_2=1, b_2=1$ and $c_2=2 a$ $\begin{aligned} & \therefore(-2-4 a)^2=\left(2 a^2+2\right)(2-a) \\ & \Rightarrow 16 a^2+4+16 a=4 a^2-2 a^3+4-2 a \\ & \Rightarrow 2 a^3+12 a^2+18 a=0 \\ & \Rightarrow 2 a\left(a^2+6 a+9\right)=0 \Rightarrow 2 a(a+3)^2=0 \\ & \therefore a=-3 \quad(\because a \neq 0) \end{aligned}$ $\therefore$ Given equation is $\begin{aligned} & -3 x^2-4 x+6=0 \Rightarrow 3 x^2+4 x-6=0 \\ & \Rightarrow x=\frac{-4 \pm \sqrt{88}}{6}=\frac{-2 \pm \sqrt{22}}{3} \therefore x=\frac{-2+\sqrt{22}}{3} \end{aligned}$
Here, $a_1=2, b_1=a, c_1=-2, a_2=1, b_2=1$ and $c_2=2 a$ $\begin{aligned} & \therefore(-2-4 a)^2=\left(2 a^2+2\right)(2-a) \\ & \Rightarrow 16 a^2+4+16 a=4 a^2-2 a^3+4-2 a \\ & \Rightarrow 2 a^3+12 a^2+18 a=0 \\ & \Rightarrow 2 a\left(a^2+6 a+9\right)=0 \Rightarrow 2 a(a+3)^2=0 \\ & \therefore a=-3 \quad(\because a \neq 0) \end{aligned}$ $\therefore$ Given equation is $\begin{aligned} & -3 x^2-4 x+6=0 \Rightarrow 3 x^2+4 x-6=0 \\ & \Rightarrow x=\frac{-4 \pm \sqrt{88}}{6}=\frac{-2 \pm \sqrt{22}}{3} \therefore x=\frac{-2+\sqrt{22}}{3} \end{aligned}$
Hence, the correct answer is option: 'D'.