Question

Find the value of k such that the polynomial $x^{2}-(k+6)x+2(2k-1)$ has sum of its zeroes equal to half their product

Answer 1

Step-by-step explanation:

Given :-

The quadratic Polynomial x^2-(k+6)x+2(2k-1) has sum of its zeroes is equal to the half of its product.

To find:-

Find the value of k?

Solution:-

Given quadratic polynomial is x^2-(k+6)x+2(2k-1)

Let P(x)=x^2-(k+6)x+2(2k-1)

On comparing with ax^2+bx+c then

a=1

b= -(k+6)

c=2(2k-1)

Sum of zeroes of quadratic polynomial = -b/a

=>-(-(k+6))/1

=>k+6

Sum of the zeroes of P(x)= k+6

Product of the zeroes of quadratic polynomial=c/a

=>2(2k-1)/1

=>2(2k-1)

Product of the zeroes = 2(2k-1)

Given that

sum of its zeroes is equal to the half of its product.

Sum of the zeroes = Product of the zeroes / 2

=>-b/a = (c/a)/2

=>k+6 = 2(2k-1)/2

=>k+6 = 2k-1

= 2k-k = 6+1

=>k=7

Therefore,k=7

Answer:-

The value of k for the given problem is 7

Check:-

If k = 7 then the quadratic polynomial becomes

x^2-(7+6)x +2[(2×7)-1]

=>x^2 -13x+2(14-1)

=>x^2 -13x +2(13)

=>x^2 -13x +26

Sum of the zeroes = -b/a = 13

Product of the zeores = c/a = 26

=>26/2

=>13

=>Sum of the zeores

=>Sum of the zeroes = Product of the zeroes /2

Verified the given relation

Used formulae:-

ax^2 + bx+c is a quadratic polynomial then

• Sum of the zeores = -b/a
• Product of the zeroes = c/a