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\frac{\left(1-\sqrt{53}\right)^{2}}{4^{2}}+\left(\frac{1+\sqrt{53}}{4}\right)^{2}
To raise \frac{1-\sqrt{53}}{4} to a power, raise both numerator and denominator to the power and then divide.
\frac{\left(1-\sqrt{53}\right)^{2}}{4^{2}}+\frac{\left(1+\sqrt{53}\right)^{2}}{4^{2}}
To raise \frac{1+\sqrt{53}}{4} to a power, raise both numerator and denominator to the power and then divide.
\frac{\left(1-\sqrt{53}\right)^{2}+\left(1+\sqrt{53}\right)^{2}}{4^{2}}
Since \frac{\left(1-\sqrt{53}\right)^{2}}{4^{2}} and \frac{\left(1+\sqrt{53}\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.
\frac{1-2\sqrt{53}+\left(\sqrt{53}\right)^{2}}{4^{2}}+\frac{\left(1+\sqrt{53}\right)^{2}}{4^{2}}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{53}\right)^{2}.
\frac{1-2\sqrt{53}+53}{4^{2}}+\frac{\left(1+\sqrt{53}\right)^{2}}{4^{2}}
The square of \sqrt{53} is 53.
\frac{54-2\sqrt{53}}{4^{2}}+\frac{\left(1+\sqrt{53}\right)^{2}}{4^{2}}
Add 1 and 53 to get 54.
\frac{54-2\sqrt{53}}{16}+\frac{\left(1+\sqrt{53}\right)^{2}}{4^{2}}
Calculate 4 to the power of 2 and get 16.
\frac{54-2\sqrt{53}}{16}+\frac{1+2\sqrt{53}+\left(\sqrt{53}\right)^{2}}{4^{2}}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{53}\right)^{2}.
\frac{54-2\sqrt{53}}{16}+\frac{1+2\sqrt{53}+53}{4^{2}}
The square of \sqrt{53} is 53.
\frac{54-2\sqrt{53}}{16}+\frac{54+2\sqrt{53}}{4^{2}}
Add 1 and 53 to get 54.
\frac{54-2\sqrt{53}}{16}+\frac{54+2\sqrt{53}}{16}
Calculate 4 to the power of 2 and get 16.
\frac{54-2\sqrt{53}+54+2\sqrt{53}}{16}
Since \frac{54-2\sqrt{53}}{16} and \frac{54+2\sqrt{53}}{16} have the same denominator, add them by adding their numerators.
\frac{108}{16}
Do the calculations in 54-2\sqrt{53}+54+2\sqrt{53}.