Question

If the Product of two of the roots of x^3+kx^2-3x+4=0 is -1 then k=

Answer 1

Step-by-step explanation:

The product of two roots is -1.

Let the third root be y.

So, y(-1)=-4 so y=4.

Since the root is 4,

${4}^{3} + k {4}^{2} - 3(4) + 4 = 0 \\ 64 + 16k - 12 + 4 = 0 \\ 16k + 56 = 0 \\ 16k = - 56 \\ k = \frac{ - 56}{16} = \frac{ - 7}{2}$

Thank you

Answer 2

Step-by-step explanation:

alpha*beta = -1

We know that

alpha * beta * gama = -d/a

-1 * gama = -4

gama = 4 -------(1)

alpha + beta + gama = -k/1 = -k

alpha + beta = -k - 4 = -(k + 4) --------(2)

alpha*beta + beta*gama + gama*alpha = -3

-1 + 4beta + 4alpha = -3

4alpha + 4beta = -2

alpha + beta = -2/4 = -1/2 -----(3)

from equation (2) & (3)

-1/2 = -k - 4

k = -4 + 1/2

k = -7/2