How do you solve $- 5 + 4 b^{2} = 31$ $-5+ 4b ^ { 2} = 31$ ?

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Answer 1 · 2018-05-18T16:37:57

$b = \pm 3$ $b = +- 3$

$- 5 + 4 b^{2} = 31$ $-5 + 4b^2 = 31$

Follow SADMEP to Solve:

SA: Subtract or Add by Opposite Action
$4 b^{2} = 31 + 5 = 36$ $4b^2 = 31 + 5 = 36$

DM: Divide or Multiply by Opposite Action
$b^{2} = \frac{36}{4}$ $b^2 = 36/4$
$b^{2} = 9$ $b^2 = 9$

E: Exponents by Opposite Action (so Square Root)
$b = \pm \sqrt{9}$ $b = +- sqrt(9)$
$b = \pm 3$ $b = +- 3$

P: Drop Paranthesis and start over at SA

How do you solve −5+4b2=31-5+ 4b ^ { 2} = 31?