Question 14

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Jambmaths question: 

if two graph $y=p{{x}^{2}}+q$ and $y=2{{x}^{2}}-1$intersect at x = 2. Find the value of p in terms of q

Option A: 

$\frac{q-8}{7}$

Option B: 

$\frac{7-q}{4}$

Option C: 

$\frac{8-q}{2}$

Option D: 

$\frac{7+q}{8}$

Jamb Maths Solution: 

$\begin{align}  & y=p{{x}^{2}}+q----(i) \\ & y=2{{x}^{2}}-1---(ii) \\ & \text{At the point of intersection} \\ & \text{ }px+q=2{{x}^{2}}-1 \\ & \text{At }x=2 \\ & p{{(2)}^{2}}+q=2{{(2)}^{2}}-1 \\ & 4p+q=7 \\ & 4p=7-q \\ & p=\frac{7-q}{4} \\\end{align}$

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