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JEE Main 20127 MayOnline
Exam
JEE Main
Year
2012
Date
7 May
Mode
Online
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2012 · 7 May · Online · Mathematics · Quadratic Equation

The value of k for which the equation (K-2) x^2+8 x+K+4=0 has both...

The value of k\mathrm{k} for which the equation (K−2)x2+8x+K+4=0(K-2) x^2+8 x+K+4=0 has both roots real, distinct and negative is

Options

  1. A6
  2. B3
  3. C4
  4. D1

Correct answer

B. 3

Step-by-step solution

(K−2)x2+8x+K+4=0(K-2) x^2+8 x+K+4=0 If real roots then, 82−4(K−2)(K+4)>0⇒K2+2K−8<16⇒K2+6K−4K−24<0⇒(K+6)(K−4)<0⇒−6<K<4\begin{aligned}& 8^2-4(K-2)(K+4)>0 \\ \Rightarrow & K²+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 αβ\alpha \beta is +ve+v e⇒K+4K−2>0⇒K>−4 Also, K−2K+4>0⇒K>2\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}Rootsarerealso,Roots are real so,-6 < K < 4So,6and4arenotcorrect.Since,So, 6 and 4 are not correct. Since,K > 2,so1isalsonotcorrectvalueof, so 1 is also not correct value ofK$. \therefore K=3
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