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# need help with figuring this problem out?

0
367
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I was helping my brother with a problem and was stumped at this problem.

The solution contains a fraction +/- another fraction with a radical numerator. What is this fraction with a radical numerator?

Solve for x:  $$2x^2+3x-8=0$$

Nov 2, 2015

#1
+22881
+35

Solve for x: $$2x^2+3x-8=0$$

$$\begin{array}{rcl} \boxed{~ \begin{array}{rcl} ax^2+bx+c &=& 0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \end{array} ~}\\ 2x^2+3x-8&=&0 \qquad a= 2 \quad b = 3 \quad c = -8\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x &=& {-3 \pm \sqrt{3^2-4\cdot 2 \cdot(-8)} \over 2\cdot 2}\\ x &=& {-3 \pm \sqrt{9+ 64} \over 4}\\ x &=& {-3 \pm \sqrt{73} \over 4}\\ \hline x_1 &=& {-3 + \sqrt{73} \over 4}\\ x_1 &=&1.3860009363\\ \hline x_2 &=& {-3 - \sqrt{73} \over 4}\\ x_2 &=& -2.8860009363 \end{array}$$

Nov 2, 2015

#1
+22881
+35

Solve for x: $$2x^2+3x-8=0$$

$$\begin{array}{rcl} \boxed{~ \begin{array}{rcl} ax^2+bx+c &=& 0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \end{array} ~}\\ 2x^2+3x-8&=&0 \qquad a= 2 \quad b = 3 \quad c = -8\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x &=& {-3 \pm \sqrt{3^2-4\cdot 2 \cdot(-8)} \over 2\cdot 2}\\ x &=& {-3 \pm \sqrt{9+ 64} \over 4}\\ x &=& {-3 \pm \sqrt{73} \over 4}\\ \hline x_1 &=& {-3 + \sqrt{73} \over 4}\\ x_1 &=&1.3860009363\\ \hline x_2 &=& {-3 - \sqrt{73} \over 4}\\ x_2 &=& -2.8860009363 \end{array}$$

heureka Nov 2, 2015
#2
+564
0

Thank you!

I forgot all about that equation .-.

Nov 2, 2015