Question

If one zero of the quadratic polynomial 6x^2 - 6x + 8 is 1/3,then the other zero is :​

Answer 1

Answer:

p(x) = 6x^2 - bx + 8

α = 1/3

β = ?

p(1/3) = 6(1/3)^2 - (1/3)b + 8

0 = 6(1/9) - (b/3) + 8

0 = (2/3) - (b/3) + 8

0 = (2 – b)/3 + 8

0 = (2 - b + 24)/3

0 = 26 - b

b = 26

⇒ p(x) = 6x^2 - 26x + 8 = 0

⇒ 0 = 6x^2 - 24x - 2x + 8

⇒ 0 = 6x(x - 4) - 2(x - 4)

⇒ 0 = (6x - 2)(x - 4)

⇒ x = 2/6, 4 = 1/3, 4

Therefore, the other zero of the polynomial p(x) is 4.

Answer 2

Step-by-step explanation:

Solution :-

Method -1:-

Given quadratic polynomial is 6x²-bx+8

On comparing with the standard quadratic polynomial ax²+bx+c then

a = 6

b = -b

c = 8

One of the zero of the polynomial = 1/3

Let the other zero be X

We know that

Sum of the zeroes = -b/a

=> (1/3)+X = -(-b)/6

=> (1+X)/3 = b/6

=> 1+X = (b/6)×3

=> 1+X = b/2

=> X = (b/2)-1

=> X = (b-2)/2

Therefore, X = (b-2)/2

Product of the zeroes = c/a

=> X(1/3) = 8/6

=> X/3 = 4/3

=> X = 4

Therefore , The other zero = 4

Method -2:-

Given quadratic polynomial is P(x) = 6x²-bx+8

One of the zero of the polynomial = 1/3

We know that

If 1/3 is zero of the polynomial then it satisfies the polynomial i.e. P(1/3) = 0

=> 6(1/3)²-b(1/3)+8 = 0

=> 6(1/9)-(b/3)+8 = 0

=> (2/3)-(b/3)+8 = 0

=> (2-b+24)/8 = 0

=> (26-b)/8 = 0

=> 26-b = 0×8

=> 26-b = 0

=> b = 26

If b = 26 then the Polynomial becomes 6x²-26x+8

To get other zero We can write it as P(x) = 0

=> 6x²-26x+8 = 0

=> 2(3x²-13x+4) = 0

=> 3x²-13x+4 = 0/2

=> 3x²-13x+4 = 0

=> 3x²-12x-x+4 = 0

=> 3x(x-4)-1(x-4) = 0

=> (x-4)(3x-1) = 0

=> x-4 = 0 or 3x-1 = 0

=> x = 4 or x = 1/3

The other zero = 4

Answer :-

The other zero of the given quadratic polynomial is 4

Used Formulae:-

♦ The standard quadratic polynomial is ax²+bx+c

♦ Sum of the zeroes = -b/a

♦ Product of the zeroes = c/a

Used Method :-

♦ Splitting the middle term