solve the system of equations. y=-2x y=x^2-8
Mathematics · Middle School · Tue Nov 03 2020
Answered on
Given the systems of equation:
y = -2x
y =x^2 - 8
Solution:
In order to solve for the systems of equation, we must equate the two equations into one single solution:
x^2 - 8 = -2x
Transpose -2x on the other side of the equation, note that when transposing a number, the sign changes.
x^2 + 2x - 8 = 0
In order to solve for the roots of an equation, we simply must look at the 2nd and 3rd value. First we must think of two numbers that when added, the answer is 2, and when multiplied, the answer is -8. Hence, if we are unable to find the number, we will use the quadratic formula. The values that satisfied the given condition are, -2 and 4
(x - 2) ( x + 4)
To check if the factors are true we will use the FOIL Method.Multiply the first term of the first equation, to the first and last term of the second equation. Then, multiply the last term of the first equation, to the first and last term of the second equation.
To clearly see how it works, here's a step by step solution.
= (x)(x)
=x^2
First term of the first equation multiplied to the first term of the second equation.
=(x)(4)
= 4x
First term of the first equation multiplied to the last term of the second equation.
=(-2)(x)
=-2x
Last term of the first equation multiplied to the first term of the second equation.
=(-2)(4)
= -8
Last term of the first equation multiplied to the fast term of the second equation.
= x^2 + 4x -2x -8
= x^2 + 2x - 8
Now since we know that the factors are correct, we can simply solve for the value of the roots.
x - 2 = 0
x = 2
x + 4 = 0
x = -4
Final answer:
x = 2
x = -4