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Sagot :
x = -3 twice
Explanation:x² + 6x + 9 = 0
Since the method for solving the question isn't specified, we will be using factorisation method to solve for x.
factors of 9 = 1, 3, 9
The two numbers when multiplied gives 9 and when added gives 6 are +3 and + 3.
using factorisation method:
x² + 3x + 3x + 9 = 0
x(x + 3) + 3(x + 3) = 0
(x + 3)(x + 3) = 0
(x+3) = 0 or (x+3) = 0
x = -3 or x = -3
x = -3 twice
The second method is because of the square practice in the question.
Using complete the square:
x² + 6x + 9 = 0
x² + 6x = -9
we half the coefficient of x and the square the result
coefficient of x = 6
1/2 of coefficient of x = 6/2
square of the result = (6/2)² = 3² = 9
Add the above result from both sides of the equation:
x² + 6x + 9 = -9 + 9
(x + 3)² = 0
square root both sides:
x + 3 = +/- √0
subtract 3 from both sides:
x +3 -3 = -3 +/-√0
x = -3 + 0 or -3 - 0
x = -3 twice
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