Question

If α, β are zeroes of x² -6x + k, what is the value of k if 3α + 2β = 20

Answer 1

${\huge{\underline{\underline{\mathbb{QUESTION:-}}}}}$

Q) If $\alpha$ and $\beta$ are the zeroes of the Equation

x² - 6x + k = 0

What is the value of k if

3$\alpha$ + 2$\beta$ = 20

${\huge{\underline{\underline{\mathbb{ANSWER\;\checkmark}}}}}$

✧ Concept -

• In these type of Questions , we simply find the relation between the zeroes .

• In this question , we will first find the relationship between the zeroes (here , sum) .

• After finding the sum of zeroes , we will solve that Equation , with the given Equation in the Question by any method .

• You must know , how to find the relation between the zeroes of the Quadratic Equation .

• Standard Quadratic Equation is in the form of :

$\boxed{ \tt \: a {x}^{2} + bx + c = 0}$

where , a ≠ 0

✧ Given :

• x² - 6x + k = 0
• $3 \alpha + 2 \beta = 20$

✧ To Find :

• The value of “k”

✧ Solution :

$\: \: \: \: \: {x}^{2} - 6x + k = 0$

here ,

• a = 1
• b = -6
• c = k

So ,

$\sf \: sum \: of \: zeroes = ( \alpha + \beta ) = \frac{ - b}{a}$

$\mapsto \rm \: ( \alpha + \beta ) = \frac{ - ( - 6)}{1}$

$\mapsto \boxed{ \: \alpha + \beta = 6}.....(i)$

& , We also know that ,

$\boxed{ 3 \alpha + 2 \beta = 20}....(ii)$

Now , solve eq(i) & eq(ii) using any method .

Let's use , substitution method .

from eq(i) ,

$\alpha + \beta = 6$

$so$

$\alpha = 6 - \beta$

Put this in eq(ii)

$\mapsto \: 3(6 - \beta ) + 2 \beta = 20$

$\mapsto \: 18 - 3 \beta + 2 \beta = 20$

$\mapsto \: - \beta = 20 - 18$

$\mapsto \: \boxed{\beta = - 2}$

put it in Equation (i)

$\mapsto \: \alpha + ( - 2) = 6$

$\mapsto \: \alpha = 6 + 2$

$\mapsto \boxed{ \alpha = 8}$

So ,

We have got both the roots , they are

• $\alpha = 8$
• $\beta = - 2$

Now ,

$\sf \: product \: of \: roots = \alpha \beta = \frac{c}{a}$

$\mapsto \: 8 \times - 2 = \frac{k}{1}$

$\mapsto \boxed{k = - 16}$

So ,

The value of k would be -16 .

_________________

Answer 2

Given :

• α and β are the zeroes of x² - 6x + k
• 3α + 2β = 20

To Find :

• The value of k

Solution :

We know that Quadratic Equation is in the form of ax² + bx + c = 0, where a ≠ 0

In given equation x² - 6x + k = 0,

a = 1

b = - 6

c = k

Now, we know that,

$\sf sum \: of \: zeroes = \dfrac{ - b}{a}$

$\sf : \implies \alpha + \beta = \dfrac{ - b}{a}$

$\sf : \implies \alpha + \beta = \frac{ - ( - 6)}{1}$

$\sf : \implies \alpha + \beta = 6 \: - ①$

Now, We are given,

$\sf 3 \alpha + 2 \beta = 20 \: - ②$

Now, From ①,

$\sf \alpha + \beta = 6$

$\sf : \implies \alpha = 6 - \beta$

Put in ②,

$\sf 3(6 - \beta ) + 2 \beta = 20$

$\sf : \implies 18 - 3 \beta + 2 \beta = 20$

$\sf : \implies 18 - \beta - 20 = 0$

$\sf : \implies - 2 - \beta = 0$

$\sf : \implies - \beta = 2$

$\sf : \implies \beta = - 2$

$\Large \boxed{\sf \beta = - 2}$

Put in ①,

$\sf \alpha + ( - 2) = 6$

$\sf : \implies \alpha - 2 = 6$

$\sf : \implies \alpha = 6 + 2$

$\sf : \implies \alpha = 8$

$\Large \boxed{\sf \alpha = 8}$

So,

• α = 8
• β = - 2

Now we know that,

$\sf \: product \: of \: roots = \dfrac{c}{a}$

$\sf : \implies \alpha \times \beta = \dfrac{c}{a}$

$\sf : \implies 8 \times - 2 = \dfrac{k}{1}$

$\sf : \implies - 16 = k$

$\sf : \implies k = - 16$

$\Large \boxed{\sf k = - 16}$

$\pink{ \bf \therefore \: The \: value \: of \: k \: is \: - 16.}$