Question

# If $a+\frac{1}{a}=\frac{17}{4}$; Find ($a-\frac{1}{a}$)

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Solution

## Step 1: Squaring both sides in the given equationGiven equation: $a+\frac{1}{a}=\frac{17}{4}$On Squaring both sides, we get,${\left(a+\frac{1}{a}\right)}^{2}={\left(\frac{17}{4}\right)}^{2}$${a}^{2}+\frac{1}{{a}^{2}}+2=\frac{289}{16}$ ( As ${\left(a+b\right)}^{2}={a}^{2}+{b}^{2}+2ab$ )${a}^{2}+\frac{1}{{a}^{2}}=\frac{289}{16}-2$ ( Transposing $2$from left side to right side and while doing so addition becomes subtraction )${a}^{2}+\frac{1}{{a}^{2}}=\frac{289-32}{16}$ ( Taking LCM of Right side part and solving further )${a}^{2}+\frac{1}{{a}^{2}}=\frac{257}{16}$----------- (i) Step 2: Now, Squaring $a-\frac{1}{a}$${\left(a-\frac{1}{a}\right)}^{2}={a}^{2}+\frac{1}{{a}^{2}}-2$ ( As ${\left(a-b\right)}^{2}={a}^{2}+{b}^{2}-2ab$)${\left(a-\frac{1}{a}\right)}^{2}=\frac{257}{16}-2$ ( Substituting the value of ${a}^{2}+\frac{1}{{a}^{2}}$from (i) )${\left(a-\frac{1}{a}\right)}^{2}=\frac{257-32}{16}$( Taking LCM of Right side part and solving further )${\left(a-\frac{1}{a}\right)}^{2}=\frac{225}{16}$Taking square root on both sides,$\sqrt{{\left(a-\frac{1}{a}\right)}^{2}}=\sqrt{\frac{225}{16}}$$\left(a-\frac{1}{a}\right)=\frac{15}{4}$Hence, the value of $\left(a-\frac{1}{a}\right)$is $\frac{15}{4}$.

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