How do you solve the equation $\frac{1}{7} x^{2} - 3 = 4$ $1/7x^2-3=4$ ?

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How do you solve the equation $\frac{1}{7} x^{2} - 3 = 4$ $1/7x^2-3=4$ ?

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Answer 1 · 2017-04-03T15:43:53

$x = \pm 7$ $x=+-7$

Explanation:

Start by adding 3 to both sides of the equation.

$1/7x^2cancel(-3)cancel(+3)=4+3$

$rArr1/7x^2=7$

multiply both sides by 7 to eliminate the fraction.

$cancel(7)^1xx1/cancel(7)^1 x^2=7xx7$

$rArrx^2=49$

Take the $color(blue)"square root of both sides"$

$sqrt(x^2)=+-sqrt49$

$rArrx=+-7larrcolor(red)" are the solutions"$

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How do you solve the equation $\frac{1}{7} x^{2} - 3 = 4$ $1/7x^2-3=4$ ?

1 Answer

Explanation:

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