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# solving this inequality?

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$${-3x^2-12≤3x^2-5x-68}$$

Jun 7, 2018
edited by Guest  Jun 7, 2018

#1
+21978
+2

solving this inequality?

$$-3x^2-12 \le 3x^2-5x-68$$

-3x^2-12 \le 3x^2-5x-68

$$\begin{array}{|rcll|} \hline -3x^2-12 &\le& 3x^2-5x-68 \\ -3x^2-12 -(3x^2-5x-68 ) &\le& 0 \\ -6x^2 + 5x + 56 &\le& 0 \\ \hline \end{array}$$

Now all we have to do is find the zeroes of $$y = -6x^2 + 5x + 56$$

$$\begin{array}{|rcll|} \hline -6x^2 + 5x + 56 &=& 0 \\ x &=& \frac{-5\pm \sqrt{25-4\cdot(-6)\cdot 56 } }{2\cdot(-6)} \\ x &=& \frac{-5\pm \sqrt{1369 } }{-12} \\ x &=& \frac{-5\pm 37 }{-12} \\\\ x_1 &=& \frac{-5+37}{-12} \\ x_1 &=& \frac{32}{-12} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{-\frac{8}{3}} \\\\ x_2 &=& \frac{-5-37}{-12} \\ x_2 &=& \frac{-42}{-12} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{\frac{7}{2}} \\ \hline \end{array}$$

Since $$y = -6x^2 + 5x + 56$$  graphs as a down parabola,
the quadratic is below the axis on the ends:

Then the solution is:
$$x \le -\frac{8}{3} \quad \text{ or } \quad x \ge \frac{7}{2}$$

Jun 7, 2018

#1
+21978
+2

solving this inequality?

$$-3x^2-12 \le 3x^2-5x-68$$

-3x^2-12 \le 3x^2-5x-68

$$\begin{array}{|rcll|} \hline -3x^2-12 &\le& 3x^2-5x-68 \\ -3x^2-12 -(3x^2-5x-68 ) &\le& 0 \\ -6x^2 + 5x + 56 &\le& 0 \\ \hline \end{array}$$

Now all we have to do is find the zeroes of $$y = -6x^2 + 5x + 56$$

$$\begin{array}{|rcll|} \hline -6x^2 + 5x + 56 &=& 0 \\ x &=& \frac{-5\pm \sqrt{25-4\cdot(-6)\cdot 56 } }{2\cdot(-6)} \\ x &=& \frac{-5\pm \sqrt{1369 } }{-12} \\ x &=& \frac{-5\pm 37 }{-12} \\\\ x_1 &=& \frac{-5+37}{-12} \\ x_1 &=& \frac{32}{-12} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{-\frac{8}{3}} \\\\ x_2 &=& \frac{-5-37}{-12} \\ x_2 &=& \frac{-42}{-12} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{\frac{7}{2}} \\ \hline \end{array}$$

Since $$y = -6x^2 + 5x + 56$$  graphs as a down parabola,
the quadratic is below the axis on the ends:

Then the solution is:
$$x \le -\frac{8}{3} \quad \text{ or } \quad x \ge \frac{7}{2}$$

heureka Jun 7, 2018