Question

if a and b are the zero of the quadratic polynomial p(x)=6x²+x-1,then find the values of a/b+b/a+2[1/a+1/b]+3ab​

Answer 1

Answer :-

$\boxed{\sf{\frac{a}{b} + \frac{b}{a} + 2 \bigg( \frac{1}{a} + \frac{1}{b} \bigg) + 3ab} = \frac{-2}{3} }$

To Find ;-

Value of -

$\implies \sf{\frac{a}{b} + \frac{b}{a} + 2 \bigg( \frac{1}{a} + \frac{1}{b} \bigg) + 3ab}$

Explanation :-

If a and b are the zeros of p(x) =6x² + x - 1

We know that ,

$\to \boxed{ \sf{sum \: of \: zeros = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2}}}} \\ \\ \therefore \\ \to \: \boxed{ \: \sf{ a + b = \frac{ - 1}{6} } \: \: \: }$

And ,

$\to \boxed{ \sf{product \: of \: zeros = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }} \\ \therefore \\ \\ \to \boxed{\sf{ab = \frac{ - 1}{6} \: \: }}$

Hence ,

$\to \sf{\frac{a}{b} + \frac{b}{a} + 2 \bigg( \frac{1}{a} + \frac{1}{b} \bigg) + 3ab} \\ \\ \to \sf{ \frac{ {a}^{2} + {b}^{2} }{ab} + \frac{2(a + b)}{ab} + 3ab} \\ \\ \to \sf{ \frac{ {(a + b)}^{2} - 2ab }{ab} + \frac{2(a + b)}{ab} + 3ab}$

substitute the above values ,

$\to \sf{\frac{ { \big(\frac{ - 1}{6} \big)}^{2} - 2 \big(\frac{ - 1}{6} \big)}{ \frac{ - 1}{6} } + \frac{2 \times \frac{( - 1)}{6} }{ \frac{ - 1}{6} } + 3 \times \frac{( - 1)}{6}} \\ \\ \to \sf{\frac{ \frac{1}{36} + \frac{1}{3} }{ \frac{-1}{6} } + \frac{2}{6} \times \frac{6}{1} - \frac{1}{2} } \\ \\ \to \sf{ -6 \bigg( \frac{1}{36} + \frac{1}{3} \bigg) + 2 - \frac{1}{2} } \\ \\ \to \sf{ \frac{-1}{6} - 2 + 2 - \frac{1}{2} } \\ \\ \to \sf{ \frac{-1-3}{6} } \\ \\ \to \frac{-4}{6} = \frac{-2}{3}$

Answer 2

$\bold{\huge{\underline{\underline{\rm{ Given :}}}}}$

a and b are the zeros of Polynomial

$</strong><strong>\</strong><strong>b</strong><strong>o</strong><strong>x</strong><strong>e</strong><strong>d</strong><strong>{</strong><strong>→</strong><strong>6 {x}^{2} + x - 1</strong><strong>}</strong><strong>$

$\bold{\huge{\underline{\underline{\rm{ To\:Find :}}}}}$

$→ \frac{a}{b} + \frac{b}{a} + 2( \frac{1}{a} + \frac{1}{b} ) + 3ab=?$

$\rule{200}{1}$

$\huge{\underline{\underline{\purple{Answer→}}}}$

$\boxed{→\frac{a}{b} + \frac{b}{a} + 2( \frac{1}{a} + \frac{1}{b} ) + 3ab = - \frac{2}{3}}$

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$\green{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

a and b are the zeros of Polynomial (Given)

We know that -

• $sum \: of \: zeros \: = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

• $product \: of \: zeros = \frac{contant \: term}{coefficient \: of \: {x}^{2} }$

In this polynomial -

• Coefficient of x² = 6

• Coefficient of x = 1

• Constant Term = - 1

Thus,

$\boxed{\red{→a + b = - \frac{1}{6}}}$

$\boxed{\blue{→ab = - \frac{1}{6}}}$

Now come to the question -

$→ \frac{a}{b} + \frac{b}{a} + 2( \frac{1}{a} + \frac{1}{b} ) + 3ab \\ \\ → \frac{ {a}^{2} + {b}^{2} }{ab} + 2( \frac{a + b}{ab} ) + 3ab$

Using the identity :-

• ${a}^{2} + {b}^{2} = ( {a + b)}^{2} - 2ab$

$→\frac{( {a + b)}^{2} - 2ab }{ab} + 2( \frac{a + b}{ab} ) + 3ab$

Now substitute the values of (a + b) and ab :-

$→\frac{ ( - \frac{ {1}}{6})^{2} - 2 \times ( - \frac{1}{6}) }{( - \frac{1}{6}) } + 2( \frac{ - \frac{1}{6} }{ - \frac{1}{6} } ) + 3 \times ( - \frac{1}{6} ) \\ \\ → \frac{ \frac{1}{36 } + \frac{1}{3} }{ - \frac{1}{6} } + 2 \times 1 - \frac{1}{2} \\ \\ →\frac{ \frac{1 + 12}{36} }{ - \frac{1}{6} } + 2 - \frac{1}{2} \\ \\ → \frac{13}{36} \times ( - 6) + 2 - \frac{1}{2} \\ \\ → - \frac{13}{6} + \frac{4 - 1}{2} \\ \\ →\frac{ - 13}{6} + \frac{3}{2} \\ \\ → \frac{ - 13 + 9}{6} \\ \\ →\frac{ - 4}{6} \\ \\ → - \frac{2}{3}$

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