How do you solve #x^2 + 5x + 3 = 9#?

1 Answer
marfre
Jul 2, 2018

#x = -6, x = 1#

Explanation:

Given: #x^2 + 5x + 3 = 9#

Solve means to find the values of #x# that make the equation true.

Put the equation in the form: #Ax^2 + Bx + C = 0# by subtraction #9# from both sides of the equation:

#x^2 + 5x -6 = 0#

Factor the equation by finding two numbers that multiply to

#ul(A*C = -6 " and sum to " B = 5)#
#-2 * 3 = -6 " "-2+3 = 1" doesn't work"#
#-6 * 1 = -6 " "-6 + 1 = -5" doesn't work"#
#6 * -1 = -6 " " 6 + -1 = 5" works"#

#x^2 + 5x -6 = 0 => " "(x+6)(x-1) = 0#

#x + 6 = 0; " " x - 1 = 0#

#x = -6, x = 1#

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