Solve for b
b=1
b=-1
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\left(b-1\right)\left(b+1\right)=0
Consider b^{2}-1. Rewrite b^{2}-1 as b^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).
b=1 b=-1
To find equation solutions, solve b-1=0 and b+1=0.
b^{2}=1
Add 1 to both sides. Anything plus zero gives itself.
b=1 b=-1
Take the square root of both sides of the equation.
b^{2}-1=0
Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.
b=\frac{0±\sqrt{0^{2}-4\left(-1\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
b=\frac{0±\sqrt{-4\left(-1\right)}}{2}
Square 0.
b=\frac{0±\sqrt{4}}{2}
Multiply -4 times -1.
b=\frac{0±2}{2}
Take the square root of 4.
b=1
Now solve the equation b=\frac{0±2}{2} when ± is plus. Divide 2 by 2.
b=-1
Now solve the equation b=\frac{0±2}{2} when ± is minus. Divide -2 by 2.
b=1 b=-1
The equation is now solved.
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