To calculate: To rationalize the numerator of the expression \frac{\sqrt{

bobbie71G 2021-10-28 Answered
To calculate:
To rationalize the numerator of the expression \(\displaystyle{\frac{{\sqrt{{{x}}}-\sqrt{{{5}}}}}{{{x}^{{{2}}}-{25}}}}\).

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Expert Answer

Arham Warner
Answered 2021-10-29 Author has 23063 answers
Calculation:
The numerator of a rational expression can be rationalized by removing the irrational term from it.
To remove the irrational term, the numerator and denominator of the expression are multiplied by the conjugate of the numerator.
Conjugate of an expression \(\displaystyle\sqrt{{{a}}}+\sqrt{{{b}}}\) is given as \(\displaystyle\sqrt{{{a}}}-\sqrt{{{b}}}\) and vice-versa.
So the conjugate of the numerator is given as \(\displaystyle\sqrt{{{x}}}+\sqrt{{{5}}}\).
Multiplying the numerator and the denominator with the conjugate.
\(\displaystyle{\frac{{\sqrt{{{x}}}-\sqrt{{{5}}}}}{{{x}^{{{2}}}-{25}}}}={\frac{{\sqrt{{{x}}}-\sqrt{{{5}}}}}{{{x}^{{{2}}}-{25}}}}\times{\frac{{\sqrt{{{x}}}+\sqrt{{{5}}}}}{{\sqrt{{{x}}}+\sqrt{{{5}}}}}}\)
\(\displaystyle{\frac{{\sqrt{{{x}}}-\sqrt{{{5}}}}}{{{x}^{{{2}}}-{25}}}}={\frac{{{\left(\sqrt{{{x}}}-\sqrt{{{5}}}\right)}{\left(\sqrt{{{x}}}+\sqrt{{{5}}}\right)}}}{{{\left({x}^{{{2}}}-{25}\right)}{\left(\sqrt{{{x}}}+\sqrt{{{5}}}\right)}}}}\)
\(\displaystyle{\frac{{\sqrt{{{x}}}-\sqrt{{{5}}}}}{{{x}^{{{2}}}-{25}}}}={\frac{{{\left(\sqrt{{{x}}}\right)}^{{{2}}}-{\left(\sqrt{{{5}}}\right)}^{{{2}}}}}{{{\left({x}^{{{2}}}-{25}\right)}{\left(\sqrt{{{x}}}+\sqrt{{{5}}}\right)}}}}{i}.{e}.{\left({a}+{b}\right)}{\left({a}-{b}\right)}={a}^{{{2}}}-{b}^{{{2}}}\)
\(\displaystyle{\frac{{\sqrt{{{x}}}-\sqrt{{{5}}}}}{{{x}^{{{2}}}-{25}}}}={\frac{{{x}-{5}}}{{{\left({x}^{{{2}}}-{25}\right)}{\left(\sqrt{{{x}}}+\sqrt{{{5}}}\right)}}}}\) (Since square and square root are opposite operations, they cancel out)
\(\displaystyle{\frac{{\sqrt{{{x}}}-\sqrt{{{5}}}}}{{{x}^{{{2}}}-{25}}}}={\frac{{{x}-{5}}}{{{\left({x}-{5}\right)}{\left({x}+{5}\right)}{\left(\sqrt{{{x}}}+\sqrt{{{5}}}\right)}}}}{i}.{e}.{\left({a}+{b}\right)}{\left({a}-{b}\right)}={a}^{{{2}}}-{b}^{{{2}}}\)
\(\displaystyle{\frac{{\sqrt{{{x}}}-\sqrt{{{5}}}}}{{{x}^{{{2}}}-{25}}}}={\frac{{{1}}}{{{\left({x}+{5}\right)}{\left(\sqrt{{{x}}}+\sqrt{{{5}}}\right)}}}}\)
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