Question

If , are the zeros of the polynomial 2x2 – 4x + 5. find the value of (i) α2 + β2 (ii) (α - β)2

Answer 1

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

Given that

$\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: {2x}^{2} - 4x + 5$

We know that,

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

$\bf\implies \: \alpha + \beta = - \: \dfrac{( - 4)}{2} = 2$

Also, we know that

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

$\bf\implies \: \alpha \beta = \dfrac{5}{2}$

Now,

Consider,

$\red{\rm :\longmapsto\: { \alpha }^{2} + { \beta }^{2}}$

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

$\rm \: = \: {(2)}^{2} - 2 \times \dfrac{5}{2}$

$\rm \: = \: 4 - 5$

$\rm \: = \: - 1$

Hence,

$\red{\boxed{ \qquad \bf{ { \alpha }^{2} + { \beta }^{2} = - 1 \qquad}}}$

Now,

Consider,

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

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

$\rm \: = \: {(2)}^{2} - 4 \times \dfrac{5}{2}$

$\rm \: = \: 4 - 10$

$\rm \: = \: - 6$

Hence,

$\blue{\boxed{ \qquad \bf{ { (\alpha - \beta ) }^{2} = - 6 \qquad}}}$

Additional Information :-

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

$\red{\boxed{ \rm{ \alpha + \beta + \gamma = - \dfrac{b}{a} }}}$

$\red{\boxed{ \rm{ \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} }}}$

$\red{\boxed{ \rm{ \alpha \beta \gamma = - \dfrac{d}{a} }}}$

Answer 2

p(x)=2x2-4x+5

So, here a=2, b=-4, c=5

a+b= - b/a

a+ b= -(-4)/2

a+b=2

and,

ab=c/a

ab= 5/2

Now,

(a+b)2= a2+b2+2ab

(2)2=. a2+b2+2×5/2

4=. a2+b2+5

4-5=a2+b2

( a2+b2= -1 )

hence,

(a-b)2= a2+b2-2ab

(a-b)2= -1-2×5/2

(a-b)2=-1-5

(a-b)2=-6