Question

If $\alpha$ and $\beta$ are the zeroes of polynomial k$x^{2}$ + 4x+ 4 and if $\alpha^{2}$ + $\beta^{2}$ =24, find the value of k

Answer 1

$= >$

given

alpha and beta are the roots of quadratic equation kx^2+4x+4

(alpha)^2+(beta) ^2 = 24

Explanation

We know,

↪(a+b)^2 = a^2 + 2ab + b^2

therefor,

↪(alpha +beta) ^2 = (alpha) ^2+2ab+(beta) ^2

↪(alpha) ^2+(beta)^2=(alpha+beta)^2-2alpha x beta

⚫As a given

(alpha )^2+ (beta)^2=24

so,

↪(alpha +beta) ^2-2(alphaxbeta)=24

........(1)

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compare the eqn kx^2+4x+4

with general quadratic equation ax^2+bx+c

so, a=k,b=4,c=4

As we know,

alpha +beta = -b/a

and.... alpha X beta = c/a

alpha +beta = -4/k

alpha X beta = 4/k

put the above values of alpha +beta and alphaxbeta in eqn (1)

↪(-4/k)^2-2(4/k)=24

↪16/k^2-8/k=24

↪[16k-8k^2]/k^3=24

↪16k-8k^2=24k^3

↪24k^3+8k^2-16k=0

devide by k to the hole eqn...

24k^2+8k-16=0

↪24k^2+24k-16k-16=0

↪24k(k+1)-16(k+1)=0

↪(24k-16)(k+1)

↪k=16/26 = 8/13 or k= -1

the values of k are k= 8/3 or k=-1