Question

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

Answer 1

Answer:

-64/25

Step-by-step explanation:

Given Equation is x² - 6x + k.

Here, a = 1, b = -6, c = k.

Let α,β are the zeroes of the equation.

(i) Sum of zeroes:

α + β = -b/a

α + β = 6.

(ii) Product of zeroes:

αβ = k.

Given Equation is 3α - 2β = 20   ---- (iii)

On Solving (i) * 3 & (iii), we get

⇒ 3α + 3β = 18

⇒ 3α - 2β = 20

   -----------------

           5β = -2.

             β = -2/5

Substitute β = 2 in (iii), we get

⇒ 3α - 2β = 20

⇒ 3α - 2(-2/5) = 20

⇒ 3α + 4/5 = 20

⇒ 3α = 20 - 4/5

⇒ 3α = 96/5

⇒ α = 96/15

⇒ α = 32/5

Substitute α = 32/5, β = -2/5 in (ii), we get

αβ

= (32/5)(-2/5)

= -64/25.

Hope it helps!

Answer 2

$\huge \underbrace{\boxed{ \bf\dag \green{ Given} \dag }}$

$\tt \alpha , \beta \: are \: zeroes \: of \: {x}^{2} - 6x + k$

$\tt3 \alpha + 2 \beta = 20$

$\huge \underbrace{\boxed{ \bf\dag \red{ FIND} \dag }}$

$\tt y = ?$

$\huge \underbrace{\boxed{ \bf\dag \pink{ Solution} \dag }}$

$\tt {x}^{2} - 6x + y \\ \tt \alpha + \beta = \frac{ - b}{a} = \frac{ \cancel - ( \cancel - 6)}{1} = 6.....(i) \\ \tt \alpha \beta = \frac{c}{a} = \frac{a}{1} = k .....(ii) \\ \tt multipy \: eq(i) \: by3 \\ \tt 3( \alpha + \beta ) = 3(6) \\ \tt 3 \alpha + 3 \beta = 18.....(iii) \\ \tt given \: eq. \: is, \: 3 \alpha + 2 \beta = 20.....(iv) \\ \tt subtracting \: eq(iv)from(iii), \\ \tt 3 \alpha + 3 \beta = 18 \\ \tt \underline{ - 3 \alpha + 2 \beta = 20} \\ \tt \underline{ \: \: \: \: \beta \: \: = \: \: \: \: \: \: \: - 2} \\ \tt from \: eq(i), - 2 + \beta = 6 \\ \tt \implies \beta = 6 + 2 = 8 \\ \tt now, \: from \: eq(ii), \\ \tt \alpha \beta = k \\ \tt \therefore k = ( - 2)(8) = - 16$

$\underbrace{\boxed{ \bf \purple{ Answer}}}$

$\boxed{ \tt value \: of \: k \: is - 16.}$