Question

if alpha and beta are zeroes of polynomial p(x)=mx^2+4x+4 such that alpha square+beta square=24,then find value of m

Answer 1

Answer

According to the question,

α and β are the zeroes of polynomial kx² + 4x + 4

a = k

-b = -4

c = 4

α + β = -b/a = -4/k

αβ = c/a = 4/k

Given => α² + β² = 24

=> (α + β)² - 2αβ = 24

=> (-4/k)² - 2×4/k = 24

=> 16/k² - 8/k = 24

=> 16 - 8k = 24k²

=> 24k² + 8k - 16 = 0

=> 3k² + k - 2 = 0

=> 3k² + 3k - 2k - 2 = 0

=> 3k(k + 1) - 2(k + 1) = 0

=> (3k - 2)(k + 1) = 0

=> k = 2/3 or -1

Hope this helps....:)

please mark as brainliest

Answer 2

$\huge\mathfrak\blue{Answer:}$

Zero of a polynomial:

• Zero of a polynomial are the values of x for which the value of given polynomial becomes zero

Given:

• We have been given a Quadratic Polynomial p(x) = mx² + 4x + 4
• α and β are the zeros of given polynomial
• Also α² + β² = 24

To Find:

• We have to find the value of m in the given Quadratic Polynomial

Solution:

We have been given given that

$\boxed{\text{p(x) = mx² + 4x + 4 }}$

Comparing it with standard form of Quadratic polynomial we get

$\large\boxed{\text{a = m and b = 4 and c = 4}}$

$\sf{ }$

$\odot \:$ Relation Between zero and Coefficient

➡️Sum of Zeros

$\boxed{\text{α + β} = \sf{-\dfrac{b}{a}}}$

Substituting the Values

$\implies \text{α + β} = \sf{-\dfrac{4}{m}}$ -----------------(1)

➡️Product of zeros

$\boxed{\text{αβ} = \sf{\dfrac{c}{a}}}$

Substituting the values

$\implies \text{αβ} = \sf{\dfrac{4}{m}}$---------------------(2)

$\sf{ }$

$\odot \:$ According to the Question :

$\implies \boxed{\text{α² + β² = 24}}$

Adding and subtracting 2αβ in LHS

$\implies \text{α² + β² + 2αβ - 2αβ = 24}$

$\implies \text{α² + β² + 2αβ- 2αβ = 24}$

$\boxed{\text{( a + b )² = a² + b² + 2ab}}$

$\implies \text{( α + β )² - 2αβ = 24}$

Substituting the Values in Equation

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

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

Taking LCM on LHS

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

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

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

$\implies \sf{6m^2 + 2m - 4 = 0}$

___________________________

We have to find two numbers such that whose product is 24 and difference is 2

Two such numbers are 6 and 4

$\implies \sf{6m^2 + 6m - 4m - 4 = 0}$

$\implies \sf{6m( m + 1 ) - 4(m + 1) = 0}$

$\implies \sf{( m + 1 )(6m - 4) = 0}$

$\sf{ }$

Either m + 1 = 0 $\implies$ m = - 1

Or 6m - 4 = 0 $\implies$ m = 2/3

_________________________

$\huge\underline{\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}}}$

$\boxed{\sf{Value \: of \: m = \dfrac{2}{3} \: or -1}}$

_________________________