Question

If α and βα and β  are zeroes of the polynomial f(x)=x2−x−kf(x)=x2−x−k , such that α−β=9α−β=9 , the value of ‘k’ will be ​

Answer 1

Answer:-

Given:-

α & β are the zeroes of x² - x - k.

By comparing the given quadratic equation to the standard form of a quadratic equation i.e., ax² + bx + c = 0 ;

Let,

• a = 1
• b = - 1
• c = - k

We know that;

Sum of the zeroes = - b/a

So,

⟹ α + β = - ( - 1)/1

⟹ α + β = 1 -- equation (1)

And,

Product of the zeroes = c/a

⟹ αβ = - k/1

⟹ αβ = - k -- equation (2)

It is also given that,

⟹ α - β = 9

Squaring both sides we get,

⟹ (α - β)² = 9²

Using (a - b)² = (a + b)² - 4ab in LHS we get,

⟹ (α + β)² - 4αβ = 81

Substitute the respective values from equations (1) & (2).

⟹ (1)² - 4(- k) = 81

⟹ 1 + 4k = 81

⟹ 4k = 81 - 1

⟹ 4k = 80

⟹ k = 80/4

⟹ k = 20

Answer 2

★GIVEN:

• α and β  are zeroes of the polynomial 

$\implies \: f(x) = x {}^{2} - x - k$

• such that α−β=9

★find:-

• value of k

★solution:-

• we know that any polynomial line EQ

$\implies \pink{ax {}^{2} + bx + c}$

• Given polynomial we see that

$\implies \red{a = 1} \\ \\ \implies \red{b = - 1} \\ \\ \implies \red{c = - k}$

we know formula of two zerous some

$\implies \star \pink{ \alpha + \beta = \frac{ - b}{ \: \: \: \: a} }$

it means

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

squaring both sides

$\implies \pink{(\alpha + \beta ) {}^{2} = 1 {}^{2} } - - - (1)$

and two zerous multiple

$\implies \red{ \alpha \beta = \frac{c}{a} }$

it means

$\implies \: \alpha \beta = \frac{ - k}{1} \\ \\ \implies \red {\alpha \beta = - k} - - - (2)$

given that

$\implies \: \ \alpha - \beta = 9 {}^{}$

squaring both sides

$\implies \: ( \alpha - \beta) {}^{2} = 9 {}^{2}$

we know that formula of

$\implies \star \pink{( a - b) {}^{2} = a {}^{2} + b {}^{2} -2 ab}$

$\implies \alpha {}^{2} + \beta {}^{2} - 2 \alpha \beta = 81$

now add both sides +2 αβ

$\implies \: \alpha {}^{2} + \beta {}^{2} + 2 \alpha \beta - 2 \alpha \beta = 81 + 2 \alpha \beta$

we know that formula of

$\implies \pink{a {}^{2} + b {}^{2} + 2ab = (a + b) {}^{2} }$

$\implies \: ( \alpha + \beta ) {}^{2} - 4 \alpha \beta = 81$

put on EQ 1 and 2 value

$\implies \: 1 - 4( - k) = 81 \\ \\ \implies \: 1 + 4k = 81 \\ \\ \implies \: 4k = 81 - 1 \\ \\ \implies \: k = \cancel{ \frac{80}{4} } \\ \\ \implies \boxed{ \red {k = 20}}$

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I hope it helps you