Consider the equation x2 + 8x = 10.
Get into ax2 + bx+ c form
Find the following:
a=
b=
c=
Show how to solve the equation by using the quadratic formula. Round solutions to the nearest tenth if needed.

Respuesta :

1)  To write the equation in the standard form [tex]ax^{2} + bx + c = 0[/tex] you need to take everything to the left side and multiply everything, if necessary, to get all whole integers:
[tex]x^{2} + 8x = 10 \\ x^{2} + 8x - 10 = 0[/tex]
This will be your standard form of the equation.
2) To find a, b, c you just need to remember that:
- a is a coefficient in front of x^2
- b is a coefficient in front of x
- c is a constant with no x.
So, in your rewritten equation [tex]x^{2} + 8x - 10 = 0[/tex] you have a = 1, b = 8, and c = -10
3) To solve the equation using quadratic formula, you need:
- find the Discriminant D, which is [tex]D = b^{2} - 4ac[/tex]
- if D < 0 there is no solution
- if D = 0 there is one solution [tex]x = - \frac{b}{2a} [/tex]
- if D > 0 there are two solutions which are
[tex]x_{1} = \frac{-b + \sqrt{D} }{2a} \\ x_{2} = \frac{-b - \sqrt{D} }{2a} [/tex]
4) Let's solve the equation:
- [tex]D = b^{2} - 4ac = (8)(8) - (4)(1)(-10) = 64 - (-40) = 104[/tex]
- 104 > 0 => there are 2 solutions
- [tex]x_{1} = \frac{-b + \sqrt{D} }{2a} = \frac{-(8) + \sqrt{104} }{(2)(1)} = \frac{-8 + \sqrt{26 * 4} }{2} = \frac{-8 + 2 \sqrt{26} }{2} = -4 + \sqrt{26} \\ x_{2} = \frac{-b - \sqrt{D} }{2a} = \frac{-(8) - \sqrt{104} }{(2)(1)} = \frac{-8 - \sqrt{26 * 4} }{2} = \frac{-8 - 2 \sqrt{26} }{2} = -4 - \sqrt{26} [/tex]
5) So, this is your solution. Good luck!