Question

If squared difference of the zeros of the quadratic polynomial x²+px+45 is equal to 144 , find the value of p. *

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: \alpha , \beta \: be \: the \: zeroes \: of \:f(x) = {x}^{2} + px + 45$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\: \alpha + \beta = - \: \dfrac{p}{1} = - \: p$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\: \alpha \beta \: = \: \dfrac{45}{1} = 45$

Now,

According to statement,

$\rm :\longmapsto\: {( \alpha - \beta )}^{2} = 144$

$\rm :\longmapsto\: {( \alpha + \beta )}^{2} - 4 \alpha \beta = 144$

$\: \: \: \: \: \: \: \: \: \: \green{ \boxed{ \because \bf \: {(x - y)}^{2} = {(x + y)}^{2} - 4xy}}$

$\rm :\longmapsto\: {( - p)}^{2} - 4 \times 45 = 144$

$\rm :\longmapsto\: { p}^{2} - 180 = 144$

$\rm :\longmapsto\: { p}^{2} = 144 + 180$

$\rm :\longmapsto\: { p}^{2} = 324$

$\bf\implies \:p \: = \: \pm \: 18$

Additional Information :-

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

$\rm :\longmapsto\: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta) }^{3} - 3\alpha \beta ( \alpha + \beta )$

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

Cubic polynomial

$\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d, \: then$

$\green{ \boxed{ \bf \: \alpha + \beta + \gamma = - \: \dfrac{b}{a}}}$

$\green{ \boxed{ \bf \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a}}}$

$\green{ \boxed{ \bf \: \alpha \beta \gamma = - \: \dfrac{d}{a}}}$