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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

Best Answer 

 #1
avatar+25274 
+35

Solve for x: \(2x^2+3x-8=0\)

 

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

 

laugh

 Nov 2, 2015
 #1
avatar+25274 
+35
Best Answer

Solve for x: \(2x^2+3x-8=0\)

 

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

 

laugh

heureka Nov 2, 2015
 #2
avatar+564 
0

Thank you! 

 

I forgot all about that equation .-.

 Nov 2, 2015

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