Q34 / 90 · JEE Main 2012 (07 May Online) · Mathematics

The value of $\mathrm{k}$ for which the equation $(K-2) x^2+8 x+K+4=0$ has both roots real, distinct and negative is

  • A
    6
  • Correct answer
    B
    3
  • C
    4
  • D
    1

Solution

$(K-2) x^2+8 x+K+4=0$ If real roots then, $$ \begin{aligned} & 8^2-4(K-2)(K+4)>0 \\ \Rightarrow & K^2+2 K-8 < 16 \\ \Rightarrow & K^2+6 K-4 K-24 < 0 \\ \Rightarrow & (K+6)(K-4) < 0 \\ \Rightarrow & -6 < K < 4 \end{aligned} $$ If both roots are negative then $\alpha \beta$ is $+v e$ $$ \begin{aligned} & \Rightarrow \frac{K+4}{K-2}>0 \Rightarrow K>-4 \\ & \text { Also, } \frac{K-2}{K+4}>0 \Rightarrow K>2 \end{aligned} $$ Roots are real so, $-6 < K < 4$ So, 6 and 4 are not correct. Since, $K>2$, so 1 is also not correct value of $K$. $$ \therefore K=3 $$

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

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