If a and b are the zeroes of the quadratic polynomial f(x)=ax^2+bx+c,t

1.	If a and b are the zeroes of the quadratic polynomial f(x)=ax^2+bx+c,then evaluate: 1/a-1/b
Answer» 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. $\sf : \implies \: \dfrac{1}{ \alpha } - \: \dfrac{1}{ \beta } = \dfrac{ \beta - \alpha }{ \alpha \beta } = \dfrac{ - ( \alpha - \beta )}{ \alpha \beta } \\ \\ \sf : \implies \: let \: we \: assume \: that \\ \\\sf : \implies \: ( \alpha - \beta ) {}^{2} = ( \alpha + \beta ) {}^{2} + 4 \alpha \beta \\ \\ \sf : \implies \: ( \alpha - \beta ) {}^{2} = ( \frac{ - b}{a} ) {}^{2} + 4( \frac{c}{a}) \\ \\ \sf : \implies \: ( \alpha - \beta ) {}^{2} = \frac{ {b}^{2} }{ {a}^{2} } + \frac{4c}{a}$ $\sf : \implies \: ( \alpha - \beta ) = \sqrt{ \dfrac{ {b}^{2} }{ {a}^{2} } + \dfrac{4c}{a} } \\ \\ \sf : \implies \: ( \alpha - \beta ) = \sqrt{ \frac{ {b}^{2} + 4ac}{ {a}^{2} } } \\ \\ \sf : \implies \: ( \alpha - \beta ) = \frac{ \sqrt{ {b}^{2} + 4ac} }{a}$ $\sf : \implies \: \dfrac{1}{ \alpha } - \dfrac{1}{ \beta } = \dfrac{ - ( \alpha - \beta )}{ \alpha \beta } \\ \\ \sf : \implies \: \dfrac{ - \dfrac{ \sqrt{ {b}^{2} + 4ac} }{a} }{ \dfrac{c}{a} } = \frac{ - \sqrt{ {b}^{2} + 4ac} }{c}$

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

Answer»

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.

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

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

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