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4x^{2}-1=0
Multiply both sides by 4.
\left(2x-1\right)\left(2x+1\right)=0
Consider 4x^{2}-1. Rewrite 4x^{2}-1 as \left(2x\right)^{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).
x=\frac{1}{2} x=-\frac{1}{2}
To find equation solutions, solve 2x-1=0 and 2x+1=0.
x^{2}=\frac{1}{4}
Add \frac{1}{4} to both sides. Anything plus zero gives itself.
x=\frac{1}{2} x=-\frac{1}{2}
Take the square root of both sides of the equation.
x^{2}-\frac{1}{4}=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.
x=\frac{0±\sqrt{0^{2}-4\left(-\frac{1}{4}\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{1}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{0±\sqrt{-4\left(-\frac{1}{4}\right)}}{2}
Square 0.
x=\frac{0±\sqrt{1}}{2}
Multiply -4 times -\frac{1}{4}.
x=\frac{0±1}{2}
Take the square root of 1.
x=\frac{1}{2}
Now solve the equation x=\frac{0±1}{2} when ± is plus. Divide 1 by 2.
x=-\frac{1}{2}
Now solve the equation x=\frac{0±1}{2} when ± is minus. Divide -1 by 2.
x=\frac{1}{2} x=-\frac{1}{2}
The equation is now solved.