Question

The product of the roots of $2x^2 - 7x + N = 0$ is $-5.$ Find $N.$

Answer 1

By the fundamental theorem of algebra, we can factorize the quadratic as

$2x^2-7x+N=2(x-r_1)(x-r_2)$

so that expanding the right hand side gives

$2x^2-7x+N=2x^2-2(r_1+r_2)x+2r_1r_2$

$\implies\begin{cases}-2(r_1+r_2)=-7\\2r_1r_2=N\end{cases}$

We're told that the product of the roots is -5, or $r_1r_2=-5$, so we get

$2(-5)=N=-10$

Answer 2

The value of N for given conditions is -10. It can be calculated as shown below.

Given that:

The quadratic equation given is:  $2x^2 -7x + N = 0$

The product of roots of given equation is: -5

To find: Value of N

Explanation for product of roots and Calculations:

For quadratic equation $ax^2 + bx + c = 0$, we have:

$\text{Product of roots} = \dfrac{c}{a}\\\\\text{Sum of roots} = \dfrac{-b}{a}$

For given quadratic equation, we have:

a = 2, b = -7, c = N

Thus, by above formula:

$\text{Product of roots} = \dfrac{c}{a} = \dfrac{N}{2} = -5\\\\N = -5 \times 2 = -10$

Thus, value of N for given conditions is -10.

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