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\left(\sqrt{5}\right)^{2}-2\sqrt{5}+1+\sqrt{20}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-1\right)^{2}.
5-2\sqrt{5}+1+\sqrt{20}
The square of \sqrt{5} is 5.
6-2\sqrt{5}+\sqrt{20}
Add 5 and 1 to get 6.
6-2\sqrt{5}+2\sqrt{5}
Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.
6
Combine -2\sqrt{5} and 2\sqrt{5} to get 0.