Question

If alpha and bita are the zero of the quadratic polynomial f(x)=ax^2+b+c,
the evaluate alpha-bita​

Answer 1

${\large{\underbrace{\underline{\sf{Question}}}}}$

Q) If alpha and beta are the zeroes of the quadratic polynomial f(x) = ax² + bx + c . Then , find alpha-beta.

${\large{\underbrace{\underline{\sf{Answer\:\checkmark}}}}}$

$\frak{Given}\begin{cases}\sf{Quadratic\;Polynomial=\bf{f(x)=ax^2+bx+c}}\\ \sf{Zeroes→\bf{ \alpha \; and \; \beta}}\end{cases}$

★ To Find :-

• $\alpha - \beta$

★ Solution :-

We know ,

$\rm \mapsto \: sum \: of \: zeroes = \alpha + \beta = \frac{ - b}{a} \: ....(i)$

&

$\mapsto \rm \: product \: of \: zeroes = \alpha \beta = \frac{c}{a} \: ....(ii)$

Squaring both sides of equation (i)

$\rightarrow \rm \: {( \alpha + \beta )}^{2} = {( \frac{ - b}{a} )}^{2}$

$\rightarrow \rm \: { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = \frac{ {b}^{2} }{ {a}^{2} }$

From eq(ii)

$\rightarrow \rm \: { \alpha }^{2} + { \beta }^{2} + 2 \frac{c}{a} = \frac{ {b}^{2} }{ {c}^{2} }$

$\rightarrow \rm \: { \alpha }^{2} + { \beta }^{2} = \frac{ {b}^{2} }{ {c}^{2} } - \frac{2c}{a}$

$\rightarrow \rm \: { \alpha }^{2} + { \beta }^{2} = \frac{ {b}^{2} - 2ac}{ {a}^{2} } \: .....(iii)$

Now,

$\rightarrow \rm \: {( \alpha - \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta$

from eq(iii) ,

$\rightarrow \rm \: {( \alpha - \beta )}^{2} = \frac{ {b}^{2} - 2ac }{ {a}^{2} } - 2 \frac{c}{a}$

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

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

$\rightarrow \rm \: \alpha - \beta = \sqrt{ \frac{ {b}^{2} - 4ac }{ {a}^{2} } }$

$\rightarrow \rm \: \underline{ \boxed{ \sf{ \pink{ \alpha - \beta = \frac{ \sqrt{ {b}^{2} - 4ac} }{a} }}}}$

So,

alpha - beta = √b²-4ac/a .