+1904 \({x}^{2}+4x=12\)
Solve by factoring, completing the square, and using the quadratic formula.
gibsonj338:
You got nothing else to do? What is the point of solving a quadratic equation 3 times??
+1904 It is to show that the answer does not have to be answered by the one way. There are more than one way to solve a problem.
+128570 x^2 + 4x = 12 subtract 12 from both sides
x^2 + 4x - 12 = 0 factor
(x + 6) (x - 2) = 0 set both factors to 0 and we get that x = -6 and x = 2
Complete the square
x^2 + 4x = 12 take 1/2 of 4 = 2......square it = 4 and add to both sides
x^2 + 4x + 4 = 12 + 4 simplify the right, factor the left
(x + 2)^2 = 16 take the pos/neg square roots of both sides
x + 2 = +/- square root (16)
x + 2 = +/- 4
We have two equations
x + 2 = 4 and x + 2 = -4
Subtract 2 from both sides
x = 2 and x = -6 [just as before]
Quadratic Formula ......put the eqatuion in standard form ..... x^2 + 4x - 12 = 0
a = 1, b = 4 and c = -12 and we have
x = [ -4 +/- square root [ 4^2 -(4)(1)(-12)] ] / [2 (1)]
x = [ -4 +/- square root [ 16 + 48] ] / 2
x = [ -4 +\- square root [ 64] ] / 2
x = [ -4 + 8] / 2 = 4 / 2 = 2
And
x = [ -4 - square root (64)] / 2 =
x = [ -4 - 8] / 2 = -12 / 2 = -6 just as before......
