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$\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\dfrac{4x-1}{2}$ .

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Hint: We have been given the equation $\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\dfrac{4x-1}{2}$ . So first rationalize the denominator, after that simplify the equation. Now square both sides you will get the answer.

Complete step-by-step answer:

In elementary algebra, root rationalization is a process by which radicals in the denominator of an algebraic fraction are eliminated.

This technique may be extended to any algebraic denominator, by multiplying the numerator and the denominator by all algebraic conjugates of the denominator, and expanding the new denominator into the norm of the old denominator. However, except in special cases, the resulting fractions may have huge numerators and denominators, and, therefore, the technique is generally used only in the above elementary cases.

In mathematics, an algebraic function is a function that can be defined as the root of a polynomial equation. Quite often algebraic functions are algebraic expressions using a finite number of terms, involving only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power.

Sometimes the denominator might be more complicated and include other numbers as well as the surd.

If this is the case, you need to multiply the fraction by a number that will cancel out the surd. Remember to multiply the numerator by the same number or you will change the value of the fraction.

Rationalizing an expression means getting rid of any surds from the bottom (denominator) of fractions.

Usually when you are asked to simplify an expression it means you should also rationalize it.

A fraction whose denominator is a surd can be simplified by making the denominator rational. This process is called rationalizing the denominator.

If the denominator has just one term that is the surd, the denominator can be rationalized by multiplying the numerator and denominator by that surd.

Now we are given $\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\dfrac{4x-1}{2}$ .

So now rationalizing the denominator in above we get,

\[\begin{align}

& \left( \dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} \right)\left( \dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}} \right)=\dfrac{4x-1}{2} \\

& \left( \dfrac{{{\left( \sqrt{x+1}+\sqrt{x-1} \right)}^{2}}}{{{\left( \sqrt{x+1} \right)}^{2}}-{{\left( \sqrt{x-1} \right)}^{2}}} \right)=\dfrac{4x-1}{2} \\

\end{align}\]

Now simplifying we get,

$\begin{align}

& \left( \dfrac{{{\left( \sqrt{x+1}+\sqrt{x-1} \right)}^{2}}}{{{\left( \sqrt{x+1} \right)}^{2}}-{{\left( \sqrt{x-1} \right)}^{2}}} \right)=\dfrac{4x-1}{2} \\

& \left( \dfrac{2x+2\sqrt{x+1}\sqrt{x-1}}{x+1-x+1} \right)=\dfrac{4x-1}{2} \\

& \left( \dfrac{2x+2\sqrt{x+1}\sqrt{x-1}}{2} \right)=\dfrac{4x-1}{2} \\

& x+\sqrt{x+1}\sqrt{x-1}=\dfrac{4x-1}{2} \\

& \sqrt{x+1}\sqrt{x-1}=\dfrac{4x-1}{2}-x \\

& \sqrt{x+1}\sqrt{x-1}=\dfrac{4x-1-2x}{2} \\

& \sqrt{x+1}\sqrt{x-1}=\dfrac{2x-1}{2} \\

& 2\sqrt{{{x}^{2}}-1}=2x-1 \\

\end{align}$

Now squaring both sides we get,

$\begin{align}

& 4({{x}^{2}}-1)={{\left( 2x-1 \right)}^{2}} \\

& 4({{x}^{2}}-1)=\left( 4{{x}^{2}}+1-4x \right) \\

& 4{{x}^{2}}-4=\left( 4{{x}^{2}}+1-4x \right) \\

& -4=1-4x \\

& 4x=1+4 \\

& x=\dfrac{5}{4} \\

\end{align}$

Therefore, we get the value of $x$ as $\dfrac{5}{4}$ .

Note: Read the question carefully. Also, take utmost care that no terms are missing. Do not make silly mistakes while solving. While simplifying, take care that you solve it step by step. Do not confuse while solving.

Complete step-by-step answer:

In elementary algebra, root rationalization is a process by which radicals in the denominator of an algebraic fraction are eliminated.

This technique may be extended to any algebraic denominator, by multiplying the numerator and the denominator by all algebraic conjugates of the denominator, and expanding the new denominator into the norm of the old denominator. However, except in special cases, the resulting fractions may have huge numerators and denominators, and, therefore, the technique is generally used only in the above elementary cases.

In mathematics, an algebraic function is a function that can be defined as the root of a polynomial equation. Quite often algebraic functions are algebraic expressions using a finite number of terms, involving only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power.

Sometimes the denominator might be more complicated and include other numbers as well as the surd.

If this is the case, you need to multiply the fraction by a number that will cancel out the surd. Remember to multiply the numerator by the same number or you will change the value of the fraction.

Rationalizing an expression means getting rid of any surds from the bottom (denominator) of fractions.

Usually when you are asked to simplify an expression it means you should also rationalize it.

A fraction whose denominator is a surd can be simplified by making the denominator rational. This process is called rationalizing the denominator.

If the denominator has just one term that is the surd, the denominator can be rationalized by multiplying the numerator and denominator by that surd.

Now we are given $\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\dfrac{4x-1}{2}$ .

So now rationalizing the denominator in above we get,

\[\begin{align}

& \left( \dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} \right)\left( \dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}} \right)=\dfrac{4x-1}{2} \\

& \left( \dfrac{{{\left( \sqrt{x+1}+\sqrt{x-1} \right)}^{2}}}{{{\left( \sqrt{x+1} \right)}^{2}}-{{\left( \sqrt{x-1} \right)}^{2}}} \right)=\dfrac{4x-1}{2} \\

\end{align}\]

Now simplifying we get,

$\begin{align}

& \left( \dfrac{{{\left( \sqrt{x+1}+\sqrt{x-1} \right)}^{2}}}{{{\left( \sqrt{x+1} \right)}^{2}}-{{\left( \sqrt{x-1} \right)}^{2}}} \right)=\dfrac{4x-1}{2} \\

& \left( \dfrac{2x+2\sqrt{x+1}\sqrt{x-1}}{x+1-x+1} \right)=\dfrac{4x-1}{2} \\

& \left( \dfrac{2x+2\sqrt{x+1}\sqrt{x-1}}{2} \right)=\dfrac{4x-1}{2} \\

& x+\sqrt{x+1}\sqrt{x-1}=\dfrac{4x-1}{2} \\

& \sqrt{x+1}\sqrt{x-1}=\dfrac{4x-1}{2}-x \\

& \sqrt{x+1}\sqrt{x-1}=\dfrac{4x-1-2x}{2} \\

& \sqrt{x+1}\sqrt{x-1}=\dfrac{2x-1}{2} \\

& 2\sqrt{{{x}^{2}}-1}=2x-1 \\

\end{align}$

Now squaring both sides we get,

$\begin{align}

& 4({{x}^{2}}-1)={{\left( 2x-1 \right)}^{2}} \\

& 4({{x}^{2}}-1)=\left( 4{{x}^{2}}+1-4x \right) \\

& 4{{x}^{2}}-4=\left( 4{{x}^{2}}+1-4x \right) \\

& -4=1-4x \\

& 4x=1+4 \\

& x=\dfrac{5}{4} \\

\end{align}$

Therefore, we get the value of $x$ as $\dfrac{5}{4}$ .

Note: Read the question carefully. Also, take utmost care that no terms are missing. Do not make silly mistakes while solving. While simplifying, take care that you solve it step by step. Do not confuse while solving.

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