Question

Find the sum of all solutions to the equation $3x(x+4) = 66$.

Answer 1

Https://photomath.net/s/o2P0wN

Answer 2

Answer:

the solutions are:

$x_{1}=-2+\sqrt{26}$

$x_{2}=-2-\sqrt{26}$

and the sum of solutions is:

-4

Step-by-step explanation:

the expression is:

$3x(x+4)=66$

doing the multiplication

$3x^2+12x=66\\\\3x^2+12x-66=0$

dividing everything by 3:

$x^2+4x-22=0$

we use the quadratic formula to find the solutions:

in this case since we have an equation in the form

$ax^2+bx+c=0$

where

$a=1\\b=4\\c=-22$

the quadratic formula is:

$x=\frac{-b+-\sqrt{b^2-4ac} }{2a} \\\\x=\frac{-4+-\sqrt{4^2-4*1*(-22))=} }{2}\\ x=\frac{-4+-\sqrt{16+88} }{2} \\ x=\frac{-4+-\sqrt{104} }{2} \\\\ x=\frac{-4+-2\sqrt{26} }{2} \\\\\\x=-4+-\sqrt{26}$

from this, using the + sign we find the first solution:

$x_{1}=-2+\sqrt{26}$

and the second solution:

$x_{2}=-2-\sqrt{26}$

the sum of the solutions is:

$x_{1}+x_{2}=-2+\sqrt{26}-2-\sqrt{26}\\ x_{1}+x_{2}=-4$