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\left(\sqrt{1}\right)^{2}-\left(\sqrt{x}\right)^{2}=1
Consider \left(\sqrt{1}-\sqrt{x}\right)\left(\sqrt{1}+\sqrt{x}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
1-\left(\sqrt{x}\right)^{2}=1
The square of \sqrt{1} is 1.
1-x=1
Calculate \sqrt{x} to the power of 2 and get x.
-x=1-1
Subtract 1 from both sides.
-x=0
Subtract 1 from 1 to get 0.
x=0
Product of two numbers is equal to 0 if at least one of them is 0. Since -1 is not equal to 0, x must be equal to 0.