Question

If a+1/a= 17/4 find the value of (a-1/a)​

Answer 1

=-17/4

explanation:

a+1/a=17/4

=-17/4

hope it's correct

Answer 2

GIVEN :-

$\\ \bullet \sf \: a + \dfrac{1}{a} = \dfrac{17}{4}$

TO FIND :-

$\\ \bullet \sf \: a - \dfrac{1}{a}$

TO KNOW :-

$\\ \boxed{ \sf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy } \\ \\ \boxed{ \sf \:(x - {y)}^{2} = {x}^{2} + {y}^{2} - 2xy}$

SOLUTION :-

We have ,

$\\ \sf \: a + \dfrac{1}{a} = \dfrac{17}{4}$

Squaring both sides , we get...

$\\ \sf \ { \left(a + \dfrac{1}{a} \right)}^{2} = { \left( \dfrac{17}{4} \right)}^{2}$

We know , (x+y)² = x² + y² + 2xy

Here ,

• x = a
• y = 1/a

Putting values , we get..

$\\ \implies\sf \: {a}^{2} + { \left( \dfrac{1}{a} \right)}^{2} + 2(a) \left( \dfrac{1}{a} \right) = \dfrac{289}{16} \\ \\ \\ \implies \sf \: {a}^{2} + { \left( \frac{1}{a} \right)}^{2} + 2 = \dfrac{289}{16} \\ \\ \\ \implies \sf \: {a}^{2} + \dfrac{1}{ { a }^{2} } = \dfrac{289}{16} - 2 \: \: \: \: \: \: - - - (1)$

Now ,

$\\ \implies \sf \: {\left( a - \dfrac{1}{a} \right)}^{2}$

We know , (x-y)² = x² + y² - 2xy

Here ,

• x = a
• y = 1/a

Putting values we get..

$\\ \implies \sf \: {\left( a - \frac{1}{a} \right)}^{2} = {a}^{2} + \dfrac{1}{ {a}^{2} } - 2(a) \left( \dfrac{1}{a} \right) \\ \\ \\ \implies \sf \: {\left( a - \frac{1}{a} \right)}^{2} = \left( {a}^{2} + \dfrac{1}{ {a}^{2} } \right) - 2$

From equation (1) , a² + 1/a² = 289/16 - 2

Putting values we get..

$\\ \implies \sf \: {\left( a - \dfrac{1}{a} \right)}^{2} = \dfrac{289}{16} - 2 - 2 \\ \\ \\ \implies \sf \: {\left( a - \dfrac{1}{a} \right)}^{2} = \dfrac{289}{16} - 4 \\ \\ \\ \implies \sf \: {\left( a - \dfrac{1}{a} \right)}^{2} = \dfrac{289 - 4(16)}{16} \\ \\ \\ \implies \sf \:{\left( a - \dfrac{1}{a} \right)}^{2} = \dfrac{289 - 64}{16} \\ \\ \\ \implies \sf \: {\left( a - \dfrac{1}{a} \right)}^{2} = \dfrac{225}{16} \\ \\ \\ \implies \sf \: a - \dfrac{1}{a} = \sqrt{ \dfrac{225}{16} } \\ \\ \\ \implies \underbrace{ \boxed{ \sf \: a - \dfrac{1}{a} = \dfrac{15}{4} }}$