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Dmitrij [34]
2 years ago
9

100 points PLEASE ANSWER ASAP thank you so much Please show work

Mathematics
2 answers:
vivado [14]2 years ago
5 0

Answer:

The mistake is that the equation on both sides is multiplied by 8. Both sides should be divided by 8.

Step-by-step explanation:

8x^2-x-1=0
8x^2-x=1   Divide both sides of the equation by 8
x^2-\frac{1}{8} x=\frac{1}{8}
x^2-\frac{1}{8} x+?=\frac{1}{8} +? Add the same value to both side
x^2-\frac{1}{8} x+\frac{1}{256} =\frac{1}{8} +\frac{1}{256}              
Use a^2-2ab+b^2=(a-b)^2
(x-\frac{1}{16} )^2 = \frac{33}{256}
x-\frac{1}{16}=+-\frac{\sqrt{33} }{16}
x_{1} =\frac{-\sqrt{33}+1 }{16}
x_{2} =\frac{\sqrt{33}+1 }{16}


shutvik [7]2 years ago
4 0

Answer:

x=\dfrac{1 \pm \sqrt{33}}{16}

Step-by-step explanation:

Given:

8x^2-x-1=0

Add 1 to both sides:

\implies 8x^2-x=1

Factor out the 8 from the left side:

\implies 8\left(x^2-\dfrac{1}{8}x\right)=1

Divide both sides by 8:

\implies x^2-\dfrac{1}{8}x=\dfrac{1}{8}

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

\implies x^2-\dfrac{1}{8}x+\dfrac{1}{256}=\dfrac{1}{8}+\dfrac{1}{256}

Factor the left side and simplifyy the right:

\implies \left(x-\dfrac{1}{16}\right)^2=\dfrac{33}{256}

Square root both sides:

\implies x-\dfrac{1}{16}=\pm\sqrt{\dfrac{33}{256}}

Simplify:

\implies x-\dfrac{1}{16}=\pm\dfrac{\sqrt{33}}{16}

Add 1/16 to both sides:

\implies x=\pm\dfrac{\sqrt{33}}{16}+\dfrac{1}{16}

Simplify:

\implies x=\dfrac{1 \pm \sqrt{33}}{16}

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