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For what positive vavlue of n are the ze...

For what positive vavlue of n are the zeros of `p(x) = 5x^(2) + nx + 12 ` in ratio 2:3 ?

A

0.42

B

1.32

C

4.56

D

15.8

Text Solution

Verified by Experts

The correct Answer is:
D

If the zeroes of P(x) are in the ratio 2:3 they must take the form 2k and 3k for some value k, and `(x-2k)(x-3k)=x^(2)-5k+6k^(2)=0` . Dividing P(x) by 5 and equating coefficients yields `(n)/(5)=-5k` and `(12)/(5)=6k^(2)` . Therefore `, k=pmsqrt((2)/(5))` . Since the problem asks for a positive value of n, we use `k=-sqrt((2)/(5))`, so n=-25k `~~15.8`
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