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\left(\sqrt{2}x\right)^{2}-9=2x\left(x-3\right)
Consider \left(\sqrt{2}x-3\right)\left(\sqrt{2}x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.
\left(\sqrt{2}\right)^{2}x^{2}-9=2x\left(x-3\right)
Expand \left(\sqrt{2}x\right)^{2}.
2x^{2}-9=2x\left(x-3\right)
The square of \sqrt{2} is 2.
2x^{2}-9=2x^{2}-6x
Use the distributive property to multiply 2x by x-3.
2x^{2}-9-2x^{2}=-6x
Subtract 2x^{2} from both sides.
-9=-6x
Combine 2x^{2} and -2x^{2} to get 0.
-6x=-9
Swap sides so that all variable terms are on the left hand side.
x=\frac{-9}{-6}
Divide both sides by -6.
x=\frac{3}{2}
Reduce the fraction \frac{-9}{-6} to lowest terms by extracting and canceling out -3.