(D) all of these
14. Solution of the differential equation p2-8p+15
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p² +8p+15
= p²+5p+3p+15
= p(p+5)+3(p+5)
= (p+5)(p+3)
p² +8p+15 will result zero , when
(p+5)(p+3)=0
=> p+5 =0. or. p+3 = 0
=> p=-5. => p=-3
So , the zeros of the quadratic polynomial are -5 and -3
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