Question

One zero of polynomial 2x2 - 5x - (2k + 1) is double than second zero. Find out both zeroes and K. ​

Answer 1

Solution

Given :-

• Polynomial equation is , 2x² - 5x - (2k+1)
• first zeros is double than second

Find :-

• Value of both zeros
• Value of k.

Explanation

Let,

second zero be x ,

than,

first zeros will be 2x.

Using Formula

★ Sum of zeros = -(coefficient of x )/(coefficient of x²)

★ products of zeros = (constant part )/(coefficient of x²)

So, Now calculate

==> Sum of zeros = -(-5)/2

==> x + 2x = 5/2

==> 3x = 5/2

==> x = 5/6

And,

==> First zeros will be = 2x = 5×2/6

==> First zeros will be = 5/3 .

_________________________

Now, calculate value of k.

==> products of zeros = -(2k+1)/2

==> 5/6 + 5/3 = -(2k+1)/2

==>( 5 + 10)/6 = -(2k+1)/2

==> 15 = -6k - 3

==> 6k = -15 - 3

==> 6k = -18

==> k = -18/6

==> k = -3

Hence

• Value of k will be = -3
• zeros of polynomial will be 5/6 & 5/3

____________________

Answer 2

Solution:-

Let, p(x)=2x²-5x-(2k+1)--------(1)

Also let, α be one root of (1).

Then the other root will be 2α.

α+2α=-(-5)/2

or, 3α=5/2

or, α=5/6

Again, α×2α=-(2k+1)/2

or, 2α²=-(2k+1)/2

or, -(2k+1)/2=2×(5/6)²

or, -(2k+1)=4×25/36

or, -2k-1=25/9

or, -2k=25/9+1

or, -2k=34/9

or, k=-34/9×1/2

or, k=-17/9 Ans.