Question

If α, β are the zeroes of quadratic polynomial x2-p(x+2)-c, then prove that (α+2) (β+2)- 4 +c = 0

Answer 1

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

The given polynomial is

$\rm :\longmapsto\:Let \: f(x) = {x}^{2} - p(x + 2) - c$

can be rewritten as

$\rm :\longmapsto\: f(x) = {x}^{2} - px - 2p - c$

$\rm :\longmapsto\: f(x) = {x}^{2} - px - ( 2p + c )$

Further it is given that

$\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: f(x).$

We know,

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

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

Also,

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

$\rm :\implies\: \alpha\beta = \dfrac{ - (2p + c)}{1} = - 2p - c$

Consider,

$\rm :\longmapsto\:( \alpha + 2)( \beta + 2) - 4 + c$

$\rm \:= \: \: \alpha \beta + 2 \beta + 2 \alpha +\cancel 4 -\cancel 4 + c$

$\rm \:= \: \: - 2p -\cancel c + 2( \alpha + \beta) + \cancel c$

$\rm \:= \: \: - \cancel{2p} \: + \: \cancel {2p}$

$\rm \:= \: \:0$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

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

$\green{ \boxed{ \bf \: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \: }}$

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