Question

3) Ifαandβarezeroesofp(x)=2x2+4kx+4,suchthatα2 +β2 = 12, find k.​

Answer 1

Answer:

p(x)= 2x2+4kx+4

a2+b2=12

So,

a+b= -b/a

a+b=-4k/2

a+b=-2k

and,

ab= c/a

ab= 4/2

ab=2

Now,

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

(-2k)2= 12+2×2

4k2=. 12+4

4k2=16

K2= 16/4

K2= 4

k=√4

(k= 2)

Answer 2

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

Given that,

$\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: p(x) = {2x}^{2} + 4kx + 4$

We know,

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

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

Also, we know that

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

$\bf\implies \: \alpha \beta = \dfrac{4}{2} = 2$

According to statement,

$\rm :\longmapsto\: { \alpha }^{2} + { \beta }^{2} = 12$

$\rm :\longmapsto\:\: {( \alpha + \beta )}^{2} - 2 \alpha \beta = 12$

Now, on substituting the values, we get

$\rm :\longmapsto\: {( - 2k)}^{2} - 2 \times 2 = 12$

$\rm :\longmapsto\: {4k}^{2} - 4 = 12$

$\rm :\longmapsto\: {4k}^{2} = 12 + 4$

$\rm :\longmapsto\: {4k}^{2} = 16$

$\rm :\longmapsto\: {k}^{2} = 4$

$\bf\implies \:k \: = \: \pm \: 2$

Additional Information :-

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

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

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

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