Question

If a and b are the zeroes of the quadratic polynomial f(x)=ax^2+bx+c,then evaluate: 1/a-1/b

Answer 1

EXPLANATION.

→ a and b are the zeroes of the quadratic

polynomial = ax² + bx + c.

→ To find 1/a - 1/b.

→ Sum of zeroes of quadratic equation.

→ a + b = -b/a.

→ products of zeroes of quadratic equation.

→ ab = c/a.

$\sf : \implies \: \dfrac{1}{ \alpha } - \: \dfrac{1}{ \beta } = \dfrac{ \beta - \alpha }{ \alpha \beta } = \dfrac{ - ( \alpha - \beta )}{ \alpha \beta } \\ \\ \sf : \implies \: let \: we \: assume \: that \\ \\\sf : \implies \: ( \alpha - \beta ) {}^{2} = ( \alpha + \beta ) {}^{2} + 4 \alpha \beta \\ \\ \sf : \implies \: ( \alpha - \beta ) {}^{2} = ( \frac{ - b}{a} ) {}^{2} + 4( \frac{c}{a}) \\ \\ \sf : \implies \: ( \alpha - \beta ) {}^{2} = \frac{ {b}^{2} }{ {a}^{2} } + \frac{4c}{a}$

$\sf : \implies \: ( \alpha - \beta ) = \sqrt{ \dfrac{ {b}^{2} }{ {a}^{2} } + \dfrac{4c}{a} } \\ \\ \sf : \implies \: ( \alpha - \beta ) = \sqrt{ \frac{ {b}^{2} + 4ac}{ {a}^{2} } } \\ \\ \sf : \implies \: ( \alpha - \beta ) = \frac{ \sqrt{ {b}^{2} + 4ac} }{a}$

$\sf : \implies \: \dfrac{1}{ \alpha } - \dfrac{1}{ \beta } = \dfrac{ - ( \alpha - \beta )}{ \alpha \beta } \\ \\ \sf : \implies \: \dfrac{ - \dfrac{ \sqrt{ {b}^{2} + 4ac} }{a} }{ \dfrac{c}{a} } = \frac{ - \sqrt{ {b}^{2} + 4ac} }{c}$

Answer 2

Given:

A quadratic equation $ax^{2} +bx+c$ .

To find:

$\frac{1}{\alpha } -\frac{1}{\beta }$  where $\alpha$ and $\beta$ are the zeroes of the quadratic equation.

Solution:

We know, for a quadratic equation $ax^{2} +bx+c$ , having zeroes $\alpha$ and $\beta$,

The sum of its zeroes $\alpha+ \beta=\frac{-b}{a}$    $(i)$   $;$ and

Product of its zeroes $\alpha \beta =\frac{c}{a}$            $(ii)$

Now,

Simplifying, the equation we get,

$\frac{1}{\alpha }- \frac{1}{\beta } =\frac{\beta -\alpha }{\alpha \beta }$    $or$   $\frac{-(\alpha -\beta )}{\alpha \beta }$     $(iii)$

We know a relation,

$(\alpha -\beta )^{2} =(\alpha +\beta )^{2} -4\alpha \beta$

Substituting the given values in the equation, we get

$(\alpha -\beta )^{2} =(\frac{-b}{a} )^{2} -4(\frac{c}{a} )$

$(\alpha -\beta )^{2} = \frac{b^{2} }{a^{2} } -4\frac{c}{a}$

$(\alpha -\beta )^{2} =\frac{b^{2}-4ac }{a^{2} }$

Hence, $\alpha -\beta =$ ±$\frac{\sqrt{b^{2}-4ac } }{a}$

Now, substituting this value in equation $(iii)$ we get

$\frac{-(\alpha -\beta )}{\alpha \beta }=$ ± $\frac{\sqrt{b^{2}-4ac } }{a(\frac{c}{a}) }$

$\frac{-(\alpha -\beta )}{\alpha \beta }=$ ± $\frac{\sqrt{b^{2}-4ac } }{c}$

Final answer:

Hence, the value of $\frac{1}{\alpha } -\frac{1}{\beta }$ is ±$\frac{\sqrt{b^{2}-4ac } }{c}$ .