The sum of the non-real roots of (p^2+p-3)(p^ | vedclass

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(B) Given equation is $(p^2+p-3)(p^2+p-2)-12=0$. Let $y = p^2+p-2$. Then the equation becomes $(y-1)y - 12 = 0$. $y^2 - y - 12 = 0$. $(y-4)(y+3) = 0$.

Vedclass · Accepted Answer

$-1$

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(B) Given equation is $(p^2+p-3)(p^2+p-2)-12=0$. Let $y = p^2+p-2$. Then the equation becomes $(y-1)y - 12 = 0$. $y^2 - y - 12 = 0$. $(y-4)(y+3) = 0$. So,$y = 4$ or $y = -3$. Case $1$: $p^2+p-2 = 4 \Rightarrow p^2+p-6 = 0$. $(p+3)(p-2) = 0$,so $p = -3, 2$ (These are real roots). Case $2$: $p^2+p-2 = -3 \Rightarrow p^2+p+1 = 0$. For this quadratic,the discriminant $D = b^2 - 4ac = 1^2 - 4(1)(1) = -3 Thus,the roots are non-real. The sum of the roots of $p^2+p+1=0$ is given by $-\frac{b}{a} = -\frac{1}{1} = -1$.

The sum of the non-real roots of $(p^2+p-3)(p^2+p-2)-12=0$ is

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