Question

If p and q are the zeroes of the polynomial p(x) =x(square) -5x+k such that p-q=1 find the value of k

Answer 1

Answer :

The value of k is 6

Given :

The quadratic polynomial is

• x² - 5x + k
• p and q are the zeroes of the polynomial
• again, p - q = 1

To Find :

• The value of k

Formulae to be used :

Relations between the coefficients of the polynomial and the zeroes :

$\sf \star \: \: Sum \: \: of \: \: the \: \: zeroes =\dfrac{-coefficient \: \: of \: \: x}{coefficient \: \: of \: \: x^{2}}$

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

Solution :

Given the zeroes of polynomial are p and q

From the relations between the coefficients of the polynomial and zeroes :

⇒p + q = -(-5)/1

⇒ p + q = 5 ...........(1)

By question we have :

⇒ p - q = 1 .............(2)

Now Product of zeroes

⇒pq = k/1

⇒ pq = k ............(3)

Adding (1) and (2) we have ,

⇒ p + q + p - q = 5 + 1

⇒ 2p = 6

⇒ p = 3

Using the value of p in (1)

⇒3 + q = 5

⇒ q = 5 - 3

⇒q = 2

Putting the value of p and q in (3) :

⇒ (3)(2) = k

⇒ k = 6

Thus the equation we obtain is :

x² - 5x + 6

Answer 2

GIVEN:

• Quadratic equation : p(x) = x² - 5x + k
• p & q are the roots
• Such that p - q = 1

TO FIND:

• k = ?

SOLUTION:

Here

• Coefficient of x² = 1
• Coefficient of x = - 5
• Constant term = k

We know that

Sum of roots(α+β)= -(Coefficient of x)/coefficient of x²

→ p + q = - ( - 5)/1 = 5

→ p + q = 5

Assuming as equation ( 1 )

→ p - q = 1 (•.• Given)

Assuming as equation ( 2 )

Product of roots (αβ) = constant/ coefficient of x²

→ pq = k/1

→ pq = k

→ q = k/p

Assuming as equation ( 3 )

Adding equation 1 & 2

p + q = 5

+ p - q = 1

( + ) ( - ) ( + )

2p = 6

→ 2p = 6

→ p = 6/2

→ p = 3

Substituting p = 3 in equation (2)

→ 3 - q = 1

→ q = p - 1

→ q = 3 - 1

→ q = 2

Hence , p = 3 & q = 2

Taking equation ( 3 )

♦ q = k/p

♦ 2 = k/3

♦ k = 6

Hence , k = 6