+0

# Solve the following equations

0
56
5
+187

-6x^2 + 24x = 0

(4x+12)(x-6) = 0

9x^2 + 6x + 1 = 0

(x+2)^2 + 5 = 8

3x - x (x-4) = - 3x^2

-2 (x-5)(x+1) -6 1/2

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Thanks for the help!

Feb 27, 2019

#1
+21842
+2

1.
-6x^2 + 24x = 0

$$\begin{array}{|lrcll|} \hline & -6x^2 + 24x &=& 0 \\ & x(24-6x) &=& 0 \\ 1.& \mathbf{x}&\mathbf{=}& \mathbf{0} \\\\ 2.& 24-6x &=& 0 \\ & 6x &=& 24 \\ & \mathbf{x}&\mathbf{=}& \mathbf{4} \\ \hline \end{array}$$

Feb 27, 2019
#2
+21842
+2

2.
(4x+12)(x-6) = 0

$$\begin{array}{|lrcll|} \hline &(4x+12)(x-6) &=& 0 \\ 1.& 4x+12 &=& 0 \\ & 4x &=& -12 \\ & \mathbf{x}&\mathbf{=}& \mathbf{-3} \\\\ 2. & x-6 &=& 0 \\ & \mathbf{x}&\mathbf{=}& \mathbf{6} \\ \hline \end{array}$$

Feb 27, 2019
#3
+21842
+2

3.
9x^2 + 6x + 1 = 0

$$\begin{array}{|rcll|} \hline 9x^2 + 6x + 1 &=& 0 \\ (3x+1)^2 &=& 0 \\ 3x+1 &=& 0 \\ 3x &=& -1 \\ \mathbf{x}&\mathbf{=}& \mathbf{-\dfrac{1}{3}} \\ \hline \end{array}$$

Feb 27, 2019
#4
+21842
+2

4.
(x+2)^2 + 5 = 8

$$\begin{array}{|rcll|} \hline (x+2)^2 + 5 &=& 8 \\ (x+2)^2 &=& 8-5 \\ (x+2)^2 &=& 3 \\ x+2 &=& \pm\sqrt{3} \\ x &=& -2 \pm\sqrt{3} \\\\ \mathbf{x_1}&\mathbf{=}& \mathbf{-2 +\sqrt{3}} \\ \mathbf{x_2}&\mathbf{=}& \mathbf{-2 -\sqrt{3}} \\ \hline \end{array}$$

Feb 27, 2019
#5
+21842
+2

5.
3x - x (x-4) = - 3x^2

$$\begin{array}{|lrcll|} \hline & 3x - x (x-4) &=& - 3x^2 \\ & 3x - x^2+4x &=& - 3x^2 \\ & 2x^2+7x &=& 0 \\ & x(7+2x) &=& 0\\ 1.& \mathbf{x}&\mathbf{=}& \mathbf{0} \\\\ 2.& 7+2x &=& 0 \\ & 2x &=& -7 \\ & \mathbf{x}&\mathbf{=}& -\mathbf{\dfrac{7}{2}} \\ \hline \end{array}$$

Feb 27, 2019