How do you solve the quadratic $25 m^{2} - 8 = - 7$ $25m^2-8=-7$ using any method?

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Answer 1 · 2016-09-08T16:12:12

$m = \frac{1}{5} or m = - \frac{1}{5}$ $m = 1/5 or m = -1/5$

Usually you make a quadratic equation equal to 0 and then solve it by factoring, completing the square, or the formula.

However, this quadratic equation is a special case because it does not have an x-term.

Move the numbers to the right side to get:

$25 m^{2} = - 7 + 8 = 1$ $25m^2 = -7+8 = 1$

$\frac{25 m^{2}}{25} = \frac{1}{25} \leftarrow$ $(25m^2)/25 = 1/25 " "larr$ divide both sides by 25

$m^{2} = \frac{1}{25} \leftarrow$ $m^2 = 1/25 " "larr"$ find the square root of both sides

$m = \pm \sqrt{\frac{1}{25}}$ $m = +-sqrt(1/25)$

$m = \frac{1}{5} or m = - \frac{1}{5}$ $m = 1/5 or m = -1/5$

How do you solve the quadratic 25m^2-8=-725m2−8=−725m^2-8=-7 using any method?