Question

Which are the solutions of x2 = –13x – 4?

Answer 1

You first have to rearrange the equation to give x2 + 13x + 4 = 0. In this equation, a =1, b=13, and c=4. Since you can not get the roots by easily factoring the equation, you have to use the quadratic equation. The quadratic equation is x = (-b +/- sqrt (b^2-4ac))/2a. If our equation, plugging in the numbers into the equation gives the following expressions: x= (-13 + sqrt (13^2 - (4)(1)(4)))/(2 x 1) , (-13 - sqrt (13^2 - (4)(1)(4)))/(2x1). The final answers will be x = (-13/2) +( 3/2)sqrt (17) , (13/2) - (3/2)sqrt(17)

Answer 2

we have

$x^{2} =-13x-4\\ x^{2} +13x+4=0$

we know /that

Since you can not get the roots by easily factoring the equation, you have to use the quadratic equation.

The quadratic equation is

$x=(-b + /-\sqrt{(b^{2}-4ac)} )/(2a)$

in this problem

$a=1\\ b=13\\ c=4$

substitute in the equation

$x=(-13 + /-\sqrt{(13^{2}-4*1*4)} )/(2)$

$x=(-13 + /-\sqrt{(153)} )/(2)$

the roots are

$x1=(-13 + \sqrt{(153)} )/(2)$

$x2=(-13 -\sqrt{(153)} )/(2)$

$x1=(- \frac{13}{2} )+( \frac{3}{2} )* \sqrt{17}$

$x2=(- \frac{13}{2} )-( \frac{3}{2} )* \sqrt{17}$