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NDA 2011
Exam
NDA
Year
2011
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2011 · Mathematics · Quadratic Equation

What are the roots of the equation 2(y+2)^2-5(y+2)=12 ?

What are the roots of the equation 2(y+2)2−5(y+2)=12?2(y+2)^2-5(y+2)=12 ?

Options

  1. A−7/2,2-7 / 2,2
  2. B−3/2,4-3 / 2,4
  3. C−5/3,3-5 / 3,3
  4. D3/2,43 / 2,4

Correct answer

A. −7/2,2-7 / 2,2

Step-by-step solution

Let y+2=xy+2=x, equation can be written as 2x2−5x−12=02x2−8x+3x−12=02x(x−4)+3(x−4)=0(x−4)(2x+3)=0\begin{array}{r}2 x^2-5 x-12=0 \\ 2 x^2-8 x+3 x-12=0 \\ 2 x(x-4)+3(x-4)=0 \\ (x-4)(2 x+3)=0\end{array}
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