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\left(3x-2\right)\left(3x+2\right)=0
Consider 9x^{2}-4. Rewrite 9x^{2}-4 as \left(3x\right)^{2}-2^{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{2}{3} x=-\frac{2}{3}
To find equation solutions, solve 3x-2=0 and 3x+2=0.
9x^{2}=4
Add 4 to both sides. Anything plus zero gives itself.
x^{2}=\frac{4}{9}
Divide both sides by 9.
x=\frac{2}{3} x=-\frac{2}{3}
Take the square root of both sides of the equation.
9x^{2}-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\times 9\left(-4\right)}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{0±\sqrt{-4\times 9\left(-4\right)}}{2\times 9}
Square 0.
x=\frac{0±\sqrt{-36\left(-4\right)}}{2\times 9}
Multiply -4 times 9.
x=\frac{0±\sqrt{144}}{2\times 9}
Multiply -36 times -4.
x=\frac{0±12}{2\times 9}
Take the square root of 144.
x=\frac{0±12}{18}
Multiply 2 times 9.
x=\frac{2}{3}
Now solve the equation x=\frac{0±12}{18} when ± is plus. Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.
x=-\frac{2}{3}
Now solve the equation x=\frac{0±12}{18} when ± is minus. Reduce the fraction \frac{-12}{18} to lowest terms by extracting and canceling out 6.
x=\frac{2}{3} x=-\frac{2}{3}
The equation is now solved.