Question

if one zero of the polynomial 2x square-ax+64 is twice rhe others find 'a'​

Answer 1

Solution

Given :-

• Equation, 2x² - ax + 64 = 0
• one zeroes is twice of other zeroes

Find :-

• Value of a

Explanation

Let,

• First & Second zeroes be = p & q

So,

==> Sum of Zeroes = - (Coefficient of x)/(Coefficient of x²)

==> p + q = - (-a)/2

==> p + q = a/2

But, by question condition,

• p = 2q

keep Value of p

==> 2q + q = a/2

==> 3q = a/2

==> q = a/6 ---------(1)

Again,

==> product of zeroes = (Constant part)/(Coefficient of x²)

==> p . q = 64/2

==> p. q = 32

By, question condition

==> 2q . q = 32

==> q² = 32/2

==> q² = 16

==> q = √16

==> q = 4 -------(2)

Keep Value of q in equ(1)

==> 4 = a/6

==> a = 4 * 6

==> a = 24

Hence

• Value of a will be = 24

_________________

Answer 2

Answer:

24 is answer of question