Question

If a and b are the roots of the quadratic equation x²+px+12=0 with the condition a-b=1, then what is the value of p?

Answer 1

Answer:

Sum of roots = a + b = -p.

Product of roots = ab = 12.

(a - b)^2 = (a + b)^2 - 4(ab) = p^2 - 48 = 1. ==>*** p = ±7.***

Case 1: x^2 + 7x + 12 = (x+4)(x+3)=0. x = -4 or -3.

Case 2: x^2 - 7x + 12 = (x -4)(x-3)=0. x = 4 or 3

Answer 2

Answer

x²+ px + 12 = 0

so let the roots be 1 d and 3 d

We know that sum of the roots = -b/a

and product of the roots = c/a

1d+3d = - p / 1

1d * 3d = 12

= 12

d = 2

So substituting, 1(2) + 3 (2) = p

2+6 = - p

p = - 8 or +8 (since root can be both positive and negative)

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