Question

Find the value of 'k' if alpha and beta are two zeros of x^{2} -5x+k ....where (alpha -beta)^2 =1

Answer 1

Answer :

The required value of k is 6

Given :

The quadratic equation is :

• x² - 5x + k
• α and β are the zeroes of the given equation
• (α - β)² = 1

To Find :

• The value of k

Concept to be used :

The relationship between coefficients and zeroes of the polynomial :

$\sf \star \: \: Sum \: \: of \: \: the \: \: zeroes = -\dfrac{Coefficient \: \: of \: x}{ Coefficient \: \: of \: x^{2}} \\\\ \sf \star \: \: Product \: \: of \: \: the \: \: zeroes = \dfrac{constant \: \: term}{ Coefficient \: \: of \: x^{2}}$

Solution :

Since the equation is :

x² - 5x + k

So from relations mentioned above we have :

α + β = -(-5)/1

⇒α + β = 5 ..........(1)

And given

(α - β)² = 1

⇒ α - β = 1 .........(2)

Adding (1) and (2) we have :

α + β + α - β = 5 + 1

⇒2α = 6

⇒ α = 3

Again putting the value in (1) we have :

3 + β = 5

⇒ β = 5 - 3

⇒ β = 2

Now from relation of product of zeroes :

αβ = k/1

⇒ k = (3)(2) [ since α = 3 and β=2]

⇒ k = 6

Answer 2

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$\huge\tt{GIVEN:}$

• alpha and beta are two zeros of x^{2} -5x+k ....where (alpha -beta)^2 =1

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$\huge\tt{TO~FIND:}$

• the value of k

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$\huge\tt{SOLUTION:}$

THE EQUATION- x² - 5x + k

The relations between them are-

↪α + β = -(-5)/1

↪α + β = 5 _____(EQ.1)

&

↪(α - β)² = 1

↪ α - β = 1 ______(EQ.2)

_______________________________

If we compare both of them, we get

↪α + β + α - β = 5 + 1

↪2α = 6

↪α = 6/2

↪α = 3

Putting the value in equation 1,

↪3 + β = 5

↪ β = 5 - 3

↪ β = 2

The relation of products of 0-

↪αβ = k/1

↪k = (3)(2)

↪k = 6

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