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sertanlavr [38]
3 years ago
10

I really need help here ASAP​

Mathematics
1 answer:
AlexFokin [52]3 years ago
5 0

Answer:

D. x=\frac{-4-\sqrt{31}}{3} or  x=\frac{-4+\sqrt{31}}{3}

Step-by-step explanation:

The given equation is:

3x^2+8x=5

Divide through by 3;

x^2+\frac{8}{3}x=\frac{5}{3}

Add the square of half the coefficient of x to both sides.

x^2+\frac{8}{3}x+(\frac{4}{3})^2=\frac{5}{3}++(\frac{4}{3})^2

x^2+\frac{8}{3}x+\frac{16}{9}=\frac{5}{3}+\frac{16}{9}

The left hand side is now a perfect square:

(x+\frac{4}{3})^2=\frac{31}{9}

Take square root

x+\frac{4}{3}=\pm \sqrt{ \frac{31}{9}}

x=-\frac{4}{3}\pm \sqrt{ \frac{31}{9}}

x=-\frac{4}{3}\pm \frac{\sqrt{31}}{3}

D. x=\frac{-4-\sqrt{31}}{3} or  x=\frac{-4+\sqrt{31}}{3}

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