Question

If α and β are zeroes of the quadratic polynomial f(x) = k2+ 4x + 4 such that α2 + β2 = 24, find the value of k.

Answer 1

Given:

α and β are the zeroes of the quadratic polynomial f(x) = kx² + 4x + 4 .

To find out:

Find the value of k ?

Concept for quadratic polynomial:

★ Sum of zeroes ( α + β ) = -coefficient of x/coefficient of x²

★ Product of zeroes ( αβ ) = constant/coefficient of x²

Solution:

$\sf{Sum \: of \: zeroes : \alpha + \beta = \dfrac{ - 4}{k} }$

$\sf{Product \: of \: zeroes : \alpha \beta = \dfrac{4}{k} }$

Now,

$\sf{ { \alpha }^{2} + { \beta }^{2} = 24} \: \: [ \: Given \: ]$

We know that,

α² + β² = ( α + β )² - 2αβ

$\implies \sf{( \alpha + \beta ) {}^{2} - 2 \alpha \beta = 24}$

Putting the values of α + β and αβ in the above, we get

$\implies \sf{ {( - \dfrac{ 4}{k}) }^{2} - 2 \times \dfrac{4}{k} = 24}$

$\implies \sf{ \dfrac{16}{ {k}^{2} } - \dfrac{8}{k} = 24}$

$\implies \sf{ \dfrac{16 - 8k}{k {}^{2} } = 24}$

$\implies \sf{16 - 8k = 24 {k}^{2} }$

$\implies \sf{24 {k}^{2} + 8k - 16 = 0}$

$\implies \sf{3 {k}^{2} + k - 2 = 0}$

By splitting middle terms

$\implies \sf{3 {k}^{2} + 3k - 2k - 2 = 0}$

$\implies \sf{3k(k + 1) - 2(k + 1) = 0}$

$\implies \sf{(k + 1)(3k - 2) = 0}$

$\implies \sf{k + 1 = 0 \: or \: 3k - 2 = 0}$

$\implies \sf{k = - 1 \: or \: k = \dfrac{2}{3} }$

Hence, the value of k = -1 or 2/3 .

Answer 2

Given :

• α and β are zeroes of the quadratic polynomial f(x) = k2+ 4x + 4 such that α2 + β2 = 24.

To find :

• Value of k =?

Step-by-step explanation :

Since, α and β are zeroes of the quadratic polynomial f(x) = k2+ 4x + 4 then,

α + β = -b/a = - 4/k

And, α, β = c/a = 4/k

It is Given that,

➟ α² + β² = 24

Now,

➟ (α + β) ² - 2αβ = 24

➟ (-4/k)² - 2(4/k) = 24

➟ 16/k² - 8/ k = 24

➟ 16 - 8k/k² = 24

➟ 16 - 8k = 24k²

➟ 24k² + 8k - 16 = 0

➟ 8(3k² + k - 2) = 0

➟ 3k² + k - 2 = 0

➟ 3k² + 3k - 2k + 2

➟ 3k(k + 1) - 2(k + 1) = 0

➟ (3k - 2) (k + 1) = 0

➟ 3k - 2 = 0 Or k + 1 = 0

➟ K = 2/3 Or k = - 1

Therefore, Value of k = K = 2/3 Or k = -1