Quadratic Equation II

Algebra Level 2

The equation p ( p − 7 ) = 5 ( 66 ) p(p-7) = 5(66) is factored into ( p − a ) ( p + b ) (p-a)(p+b) where a a and b b are positive integers and a > b a > b . What is the value of a 2 − b 2 a^{2}-b^{2} ?


The answer is 259.

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Dev Od by dev od , 30, Malaysia

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3 solutions

Shubhendra Singh
Dec 5, 2014

Given that p ( p − 7 ) = 5 ( 66 ) p(p-7)=5 (66)

We know that 5 ( 66 ) = 22 × 15 5(66)=22\times15

p 2 − 7 p − ( 22 ) ( 15 ) = 0 p^{2}-7p-(22)(15)=0

p 2 − 22 p + 15 p − ( 22 ) ( 15 ) p^{2}-22p+15p-(22)(15)

( p − 22 ) ( p + 15 ) (p-22)(p+15)

a = 22 , b = 15 a=22 \ , \ b=15

So a 2 − b 2 = 259 a^{2}-b^{2}=259

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Good solution ya, nice one.

dev od - 6 years, 6 months ago

p(p-7)=5(66) we know that,5(66)=330 p^2-7p-330=0 p^2-22p+15p-330=0 p(p-22)+15(p-22)=0 [330=22*15] (p-22) (p+15)=0
so, a=22 , b=15 [ a and b both are positive integers] 22^2-15^2=259

0 Helpful 0 Interesting 0 Brilliant 0 Confused

try ,u can find it easy to solve

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