Question

if a^2-5a +1 =0 then find a-1/a​

Answer 1

Question :-

• Find value of $\sf a - \dfrac{1}{a}$

Answer :-

• $\sf +\sqrt{21}\:,\:-\sqrt{21}$

Explanation :-

Step 1 :-

Factorise the equation and find the value of a from the equation.

$\mapsto\sf{\ a^{2}-5a+1=0}$

Let the root of the equation be A . So, eqn is

$\mapsto\sf{\ A^{2}-5A+1=0}$

$\bf{Using\: Quadratic\:Formula}$

$\sf{\boxed{\boxed{A = \dfrac{-b\pm\sqrt{\ b^{2}-4ac}}{2a}}}}$

$\mapsto\sf{A = \dfrac{-(-5)\pm\sqrt{\ (-5)^{2}-4(1)(1)}}{2(1)}}$

$\mapsto\sf{A = \dfrac{+5\pm\sqrt{25-4}}{2}}$

$\mapsto\sf{A = \dfrac{5\pm\sqrt{21}}{2}}$

$\mapsto\sf{A = \dfrac{5+\sqrt{21}}{2}\:,\:\dfrac{5-\sqrt{21}}{2}}$

$\sf{ }$

Step 2 :-

For finding

$\sf{A-\dfrac{1}{A}}$ put value of A

$\implies\sf{A-\dfrac{1}{A}}$

For 1st Value of A (+ve)

$\implies\sf{\dfrac{5+\sqrt{21}}{2}-\dfrac{2}{5+\sqrt{21}}}$

$\implies\sf{\dfrac{\ (5+\sqrt{21})^{2}-4}{2(5+\sqrt{21})}}$

$\implies\sf{\dfrac{25+21+10\sqrt{21}-4}{10+2\sqrt{21}}}$

$\implies\sf{\dfrac{42+10\sqrt{21}}{10+2\sqrt{21}}}$

$\implies\sf{\dfrac{\cancel{2}(21+5\sqrt{21})}{\cancel{2}(5+\sqrt{21}}}$

$\implies\sf{\dfrac{ \sqrt{21}\cancel{(\sqrt{21}+5)}}{\cancel{(5+\sqrt{21})}}}$

$\sf{\boxed{\sqrt{21}}}$

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For 2nd Value of A (+ve)

$\implies\sf{\dfrac{5-\sqrt{21}}{2}-\dfrac{2}{5-\sqrt{21}}}$

$\implies\sf{\dfrac{\ (5-\sqrt{21})^{2}-4}{2(5-\sqrt{21})}}$

$\implies\sf{\dfrac{25+21-10\sqrt{21}-4}{10-2\sqrt{21}}}$

$\implies\sf{\dfrac{42-10\sqrt{21}}{10-2\sqrt{21}}}$

$\implies\sf{\dfrac{\cancel{2}(21-5\sqrt{21})}{\cancel{2}(5-\sqrt{21}}}$

$\implies\sf{\dfrac{ \sqrt{21}\cancel{(\sqrt{21}-5)}}{\cancel{(5-\sqrt{21})}}}$

$\sf{\boxed{-\sqrt{21}}}$

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In the Answer we assume a = A

Answer 2

$\sf{\huge{\bold{\pink{\bigstar{\boxed{\boxed{Question}}}}}}}$

$if \: a^2 - 5a + 1 = 0$

$find\: a - \dfrac{1}{a}$

$\sf{\huge{\bold{\pink{\bigstar{\boxed{\boxed{Solution}}}}}}}$

$\implies a^2 - 5a + 1 = 0$

• Divide whole equation by a

$\implies \dfrac{a^2}{a} - \dfrac{5a}{a} + \dfrac{1}{a} = \dfrac{0}{a}$

$\implies a - 5 + \dfrac{1}{a} = 0$

$\implies a + \dfrac{1}{a} = 5$

$\sf{\bold{\red{\boxed{(a + \dfrac{1}{a})^2 - (a - \dfrac{1}{a})^2 = 4a \dfrac{1}{a}}}}}$

$\implies 5^2 - (a - \dfrac{1}{a})^2= 4$

$\implies 25 - (a- \dfrac{1}{a})^2 = 4$

$\implies ( a - \dfrac{1}{a})^2 = 21$

$\implies (a - \dfrac{1}{a}) = \sqrt{21}$

$\sf{\bold{\red{\boxed{\dag ( a -\dfrac{1}{a} )=\sqrt{21}}}}}$

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