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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2015 · Mathematics · Quadratic Equation

If 2 p+3 q=18 and 4 p^2+4 p q-3 q^2-36=0 , then what is (2 p+q)...

If 2p+3q=182 p+3 q=18 and 4p2+4pq−3q2−36=04 p^2+4 p q-3 q^2-36=0, then what is (2p+q)(2 p+q) equal to?

Options

  1. A6
  2. B7
  3. C10
  4. D20

Correct answer

C. 10

Step-by-step solution

4p2+4pq−3q2−36=0⇒(4p2+4pq+q2)−4q2−36=0(2p+q)2=4q2+362p+3q=18p=18−3q2⇒[18−3q2×2+q]2=4q2+36⇒(18−3q+q)2⇒(18−2q)2=4q2+36⇒324+4q2−72q=4q2+3672q=324−36=288\begin{aligned} & 4 p^2+4 p q-3 q^2-36=0 \\ \Rightarrow & \left(4 p^2+4 p q+q^2\right)-4 q^2-36=0 \\ & (2 p+q)^2=4 q^2+36 \\ & 2 p+3 q=18 \\ & p=\frac{18-3 q}{2} \\ \Rightarrow & {\left[\frac{18-3 q}{2} \times 2+q\right]^2=4 q^2+36 } \\ \Rightarrow & (18-3 q+q)^2 \\ \Rightarrow & (18-2 q)^2=4 q^2+36 \\ \Rightarrow & 324+4 q^2-72 q=4 q^2+36 \\ & 72 q=324-36=288\end{aligned}\mathrm{q}=\frac{288}{72}=4PuttingthevalueofPutting the value ofq in 2 p+3 q=182p+3×4=182p=18−12p=62=3(2p+q)=2×3+4=10\begin{aligned} & 2 p+3 \times 4=18 \\ & 2 p=18-12 \\ & p=\frac{6}{2}=3 \\ & (2 p+q)=2 \times 3+4=10 \end{aligned}\therefore \quad$ Option (C) is correct.
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