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Solve the given equation:
$\dfrac{7+\sqrt{5}}{7-\sqrt{5}}-\dfrac{7-\sqrt{5}}{7+\sqrt{5}}=a+\dfrac{7}{11}b\sqrt{5}$

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 Hint: We are given $\dfrac{7+\sqrt{5}}{7-\sqrt{5}}-\dfrac{7-\sqrt{5}}{7+\sqrt{5}}=a+\dfrac{7}{11}b\sqrt{5}$ , so take LHS and simplify it. After that, compare LHS and RHS, you will get the value of $a$ and $b$ .

Complete step-by-step Solution:

In 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 fromIn elementary 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 have been given $\dfrac{7+\sqrt{5}}{7-\sqrt{5}}-\dfrac{7-\sqrt{5}}{7+\sqrt{5}}=a+\dfrac{7}{11}b\sqrt{5}$ .
So taking LHS,
LHS$=\dfrac{7+\sqrt{5}}{7-\sqrt{5}}-\dfrac{7-\sqrt{5}}{7+\sqrt{5}}$
Now simplifying we get,
LHS\[=\dfrac{{{\left( 7+\sqrt{5} \right)}^{2}}-{{\left( 7-\sqrt{5} \right)}^{2}}}{\left( 7+\sqrt{5} \right)\left( 7-\sqrt{5} \right)}\]
\[\begin{align}
  & =\dfrac{49+5+14\sqrt{5}-(49+5-14\sqrt{5})}{\left( 49-5 \right)} \\
 & =\dfrac{28\sqrt{5}}{44} \\
 & =\dfrac{7\sqrt{5}}{11} \\
\end{align}\]
We get, LHS \[=\dfrac{7\sqrt{5}}{11}\].
Comparing with RHS we get,
$a+\dfrac{7}{11}b\sqrt{5}=0+\dfrac{7\sqrt{5}}{11}$
We get $a=0$ and $b=1$.

Note: Read the question carefully. Don’t confuse yourself while solving the problem. Also, take utmost care that no terms are missing. Do not jumble. While simplifying, do not make any silly mistakes. Your concept regarding rationalization should be clear.

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