Answer:
x = -10, -2
Step-by-step explanation:
The given equation is [tex]x^2+12x=-20[/tex]
We can rewrite this equation by adding 20 to both sides as
[tex]x^2+12x+20=0[/tex]
Comparing with [tex]ax^2+bx+c=0[/tex], we get
a=1
b = 12
c = 20
Now, add [tex](\frac{b}{2})^2[/tex] to both sides of the equation.
[tex](\frac{b}{2})^2\\\\=(12/2)^2\\\\=6^2=36[/tex]
So add 36 to both sides of the equation
[tex]x^2+12x+36+20=36[/tex]
Subtract 20 to both sides
[tex]x^2+12x+36=16[/tex]
We can rewrite the left hand side of the equation as
[tex](x+6)^2=16[/tex]
Take square root both sides
[tex]\sqrt{(x+6)^2}=\pm\sqrt{16}\\\\x+6=\pm4\\\\x=4-6,-4-6\\\\x=-2,-10[/tex]
Thus, the values of x are x = -10, -2
Correct response:
The solution of the equation are; x = -2 or x = -10
The given quadratic equation is; x² + 12·x = -20
Required:
To solve the quadratic equation by completing the square
Solution:
The equation is presented in the form; x² + b·x = c
By completing the square, we have;
Comparison with the given equation gives;
[tex]\mathbf{x^2 + 12 \cdot x + \left(\dfrac{12}{2} \right)^2} = -20 + \left(\dfrac{12}{2} \right)^2[/tex]
x² + 12·x + 6² = -20 + 6² = 16 = 4²
Therefore, we have;
x² + 12·x + 6² = (x + 6)² = 4²
x + 6 = ±4
x = ±4 - 6
Verification gives;
(x + 2) × (x + 10) = x² + 12·x + 20 = 0
Which gives;
x² + 12·x = -20 (the given equation)
Therefore;
The solutions of the equation x² + 12·x = -20 found by completing the square, are;
x = -2, or x = -10
Learn more about completing the square method here:
https://brainly.com/question/3939104
