Question

if alpha and beta are zeros of x square minus x + k and 3alpa +4beta=20 then find the value of k
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Answer 1

Solution : p(x) = x² - x + k ; then

→ Sum of zeros = - Coefficient of x/Coefficient of x²

→ α + β = - (- 1)/1 = 1

→ α + β = 1 __(1)

Given : 3α + 4β = 20 __(2)

From eq(1), α = 1 - β __(3)

By substituting eq(3) in (2),

→ 3(1 - β) + 4β = 20

→ 3 - 3β + 4β = 20

→ β = 17

Substituting value in eq(1)

→ α + 17 = 1

→ α = - 16

Since we know that Product of zeros = Constant Term/Coefficient of x².

→ αβ = - 16 × 17 = k/1

→ k = - 272

Answer ⇒ - 272

Answer 2

Let the polynomial be p(x) .

Then , p(x) = x²-x+k

And ,

$3 \alpha + 4 \beta = 20$

To Find :

The value of k

Solution :

We know that the relationship between the zeroes of the polynomial and the coefficients of the polynomial :

$\alpha + \beta = \frac{ - (b)}{a} = 1$

(Equation 1)

Then , we can write it as :

$\alpha = 1 - \beta$

Then , substituting the value in the equation 1 :

$3(1 - \beta ) + 4 \beta = 20 \\ \\ 3 - 3 \beta + 4 \beta = 20 \\ \\ 3 + \beta = 20 \\ \\ \beta = 20 - 3 = 17$

Then ,

$\alpha = 1 - \beta = 1 - 17 = - 16$

But any of the zeroes in p(x) :

p(x) = x²-x+k

(17)²-17+k = 0

289 -17 + k = 0

272 + k = 0

k = -272