Question

If (x – 1/x) = 15 , then find the value of (x2+ 1/x2)​

Answer 1

Question

If $(\frac{x-1}{x} )=15$, then find the value of $(x^{2}+\frac{1}{x^{2}} )$

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Answer

To find the value of ⇒ $(x^{2}+\frac{1}{x^{2}} )$ we must find the value of 'x' in this equation ⇒ $(\frac{x-1}{x} )=15$

Let's solve your equation step-by-step.

$(\frac{x-1}{x} )=15$

Step 1: Multiply both sides by x.

⇒  $\frac{x-1}{x} \times x =15\times x$

⇒ $x-1=15x$

Step 2: Subtract 15x from both sides

⇒ $x-1-15x=15x-15x$

⇒ $-14x-1=0$

Step 3: Add 1 to both sides

⇒ $-14x-1+1=0+1$

⇒ $-14x=1$

Step 4: Divide both sides by -14

⇒ $\frac{-14x}{-14} =\frac{1}{-14}$

⇒ $x=\frac{-1}{14}$

Step 5: Replace 'x' by the value to verify.

⇒ $\frac{x-1}{x} =15$

⇒ $(\frac{-1}{14}-1) \div \frac{-1}{14}$

⇒ $(\frac{-1}{14}-\frac{14}{14} ) \div \frac{-1}{14}$

⇒ $\frac{-15}{14}\div \frac{-1}{14}$

⇒ $\frac{-15}{14}\times \frac{-14}{1}$

⇒ $\frac{210}{14}$

⇒ $15$

$\bf \therefore x= \frac{-1}{14} \ in\ the \ equation => \ \frac{x-1}{x}=15$

Now, using the value of 'x' we will solve this ⇒ $(x^{2}+\frac{1}{x^{2}} )$

⇒ $(x^{2}+\frac{1}{x^{2}} )$

⇒ $((\frac{-1}{14}) ^{2}+\frac{1}{(\frac{-1}{14}) ^{2}} )$

⇒ $\frac{1}{196} +\frac{1}{\frac{1}{196} }$

⇒ $\frac{1}{196} +1\times 196$

⇒ $\frac{1}{196} + 196$

⇒ $\frac{1}{196} + \frac{38416}{196}$

⇒ $\frac{38417}{196}$

$\bf \therefore If \ (\frac{x-1}{x} )=15 \ then \ the \ value \ of \ (x^{2}+\frac{1}{x^{2}} ) \ would \ be \ \frac{38417}{196}.$

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Answer 2

$\bf \huge \underline \pink{ \mathfrak{Solution}}$


$\sf \large \bigg(x - \frac{1}{x} { \bigg)}^{2} = \bigg(15 { \bigg)}^{2} \\ \\ \\ \\ \\ \sf \large \bigg(a - b { \bigg)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ \\ \\ \\ \\ \sf \large {x}^{2} + \frac{1}{ {x}^{2} } - 2 \bigg( \cancel{x } \bigg) \bigg( \frac{1}{ \cancel{x}} \bigg) = 225 \\ \\ \\ \\ \\ \color{red} \underline{ \boxed{\sf \large {x}^{2} + \frac{1}{ {x}^{2} } = 225 + 2 - 227}}$