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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2019 · Mathematics · Quadratic Equation

What is the value of k for which the sum of the squares of the...

What is the value of kk for which the sum of the squares of the roots of 2x2−2(k−2)x−(k+1)=02 x^2-2(k-2) x-(k+1)=0 is minimum?

Options

  1. A-1
  2. B1
  3. C32\frac{3}{2}
  4. D2

Correct answer

C. 32\frac{3}{2}

Step-by-step solution

For 2x2−2(k−2)x−(k+1)=02 x^2-2(k-2) x-(k+1)=0 Let roots be α\alpha and β\beta. ∴α+β=k−2;αβ=−k+12∵α2+β2=(α+β)2−2αβ⇒α2+β2=(k−2)2+(k+1)⇒α2+β2=k2−4k+4+k+1⇒α2+β2=k2−3k+5⇒α2+β2=(k−32)2+(5−94)\begin{aligned} & \therefore \alpha+\beta=k-2 ; \alpha \beta=-\frac{k+1}{2} \\ & \because \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\ & \Rightarrow \alpha^2+\beta^2=(k-2)^2+(k+1) \\ & \Rightarrow \alpha^2+\beta^2=k^2-4 k+4+k+1 \\ & \Rightarrow \alpha^2+\beta^2=k^2-3 k+5 \\ & \Rightarrow \alpha^2+\beta^2=\left(k-\frac{3}{2}\right)^2+\left(5-\frac{9}{4}\right) \end{aligned} for minimum value k=32k=\frac{3}{2}
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