Q55 / 160 · TS EAMCET 2008 · Mathematics

If $\alpha+\beta=-2$ and $\alpha^3+\beta^3=-56$, then the quadratic equation whose roots are $\alpha$ and $\beta$ is

  • A
    $x^2+2 x-16=0$
  • B
    $x^2+2 x+15=0$
  • C
    $x^2+2 x-12=0$
  • Correct answer
    D
    $x^2+2 x-8=0$

Solution

Given that, $\alpha+\beta=-2$ and $\alpha^3+\beta^3=-56$ $$ \begin{array}{rlrl} \Rightarrow & & (\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right) & =-56 \\ \Rightarrow & \alpha^2+\beta^2-\alpha \beta & =28 \end{array} $$ $$ \begin{aligned} \text { Now, } & & (\alpha+\beta)^2 & =(-2)^2 \\ \Rightarrow & & \alpha^2+\beta^2+2 \alpha \beta & =4 \\ \Rightarrow & & 28+3 \alpha \beta & =4 \\ \Rightarrow & & \alpha \beta & =-8 \end{aligned} $$ $\therefore$ Required equation is $$ \begin{array}{rlrl} & & x^2-(-2) x+(-8) & =0 \\ \Rightarrow & x^2+2 x-8 & =0 \end{array} $$

Hence, the correct answer is option: 'D'.

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