p and q are the zeros of the quadratic
polynomial f(x) = kx² + 4x + 4 such that
p² + q² = 24 find the values of k.
Answers
★ Given -:
p and q are zeroes of f(x) = kx² + 4x + 4..
p² + q² = 24
★ To find -:
The value of k
★ We know that ,
» The sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient.
In our case,
→ p + q = - 4/ k
» The product of the roots of a quadratic equation is equal to the constant term (the third term), divided by the leading coefficient.
In our case ,
→ pq = 4 /k
» (a+b)² = a² + b² + 2ab
★ Solution -:
p+ q = -4/k
pq = 4/k
p² + q² = 24
by putting the values
\left(\dfrac{-4}{k}\right)^2 = 24 + 2 \times \dfrac{4}{k}(
k
−4
)
2
=24+2×
k
4
\dfrac{16}{k^2} = 24 + \dfrac{8}{k}
k
2
16
=24+
k
8
\dfrac{16}{k^2} - \dfrac{8}{k} = 24
k
2
16
−
k
8
=24
\dfrac{16 - 8k}{k^2} = 24
k
2
16−8k
=24
by cross multiplication
16 - 8k = 24 k^216−8k=24k
2
24 k² + 8k - 16 = 0
8(3k² + k - 2) = 0
3k² + k - 2 = 0
by middle term splitting
3k² + 3k - 2k - 2 = 0
3k (k + 1) - 2(k + 1) = 0
(3k - 2) (k + 1) = 0
3k - 2 = 0
k = 2/3
k + 1 = 0
k = -1
Therefore ,
\purple {The \: value \: of \: k = -1 \: or \: \dfrac{2}{3}}Thevalueofk=−1or
3
2