Question

The sum of the roots of a quadratic equation is -8. If one of the roots is -10, how would you determine the equation? Write the equation and explain.

Answer 1

Step-by-step explanation:

$Let \: the \: two \: root \: of \: quadratic \: eqⁿ \: to \: be \: \alpha \: \: and \: \beta . \\ \implies \alpha = - 10 \\ Now \: we \: know \: that, \: \\ \implies \alpha + \beta = - 8 \\ \implies - 10 + \beta = - 8 \\ \implies\beta = - 8 + 10 \\ \implies\beta = 2 \\ Therefore \: the \: required \: equation, \: \\ {x}^{2} - ( \alpha + \beta )x + \alpha \beta = {x}^{2} - ( - 8)x + ( - 10 \times 2) \\ \implies= {x}^{2} + 8x - 20$

Answer 2

Given:

Sum of the roots of a quadratic equation= -8

One root is= -10

To find:

The quadratic equation.

Solution:

Let the two roots be $\alpha$ and $\beta$.

$\alpha =-10$°

Now,

$\alpha +\beta =-8$

⇒$-10+\beta =-8$

⇒$\beta =-8+10$

⇒$\beta =2$

We know that the general form of the quadratic equation is:

$x^2-(\alpha +\beta)x+\alpha \beta =0$

So putting the values we get,

$x^2-(-8)x+(-10$×$2)=0$

⇒$x^2+8x-20=0$

Hence, the required equation is $x^2+8x-20=0$.