Math, asked by aswatha66, 1 month ago

10. In polynomial X^2 - kx + 6 The zeroes of are in the ratio 3 : 2 , find k.

Answers

Answered by tennetiraj86
0

Step-by-step explanation:

Given :-

In polynomial X^2 - kx + 6 The zeroes of are in the ratio 3 : 2.

To find :-

Find the value of k ?

Solution :-

Given quadratic polynomial X²-kX+6

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

We have ,

a = 1

b = -k

c = 6

Ratio of the zeroes of the given Polynomial = 3:2

Let they be 3A and 2A

We know that

The sum of the zeores = -b/a

=> 3A + 2A = -(-k)/1

=> 5A = k/1

=> 5A = k

=> k = 5A -------(1)

The Product of the zeroes = c/a

=> (3A)(2A) = 6/1

=> 6A² = 6

=> A² = 6/6

=> A² = 1

=> A = ±√1

=> A =± 1

=> A = 1 or -1 ----------(2)

On Substituting the value of A in (1) then

If A = 1 then

=> k = 5(1)

=> k = 5

If A = -1 then

=> k = 5(-1)

=> k = -5

Therefore,k = 5 or -5

Answer:-

The value of k for the given problem is 5 or -5

Check:-

If k = 5 then the Polynomial will be x²-5x+6

=> x²-5x+6 = 0

=> x²-2x-3x+6 = 0

=> x(x-2)-3(x-2) = 0

=> (x-2)(x-3) = 0

=> x-2 = 0 or x-3 = 0

=> x = 2 or 3

Zeroes are 2 and 3

Their ratio = 3:2

If k = -5 then the Polynomial will be x²+5x+6

=> x²+5x+6 = 0

=> x²+2x+3x+6 = 0

=> x(x+2)+3(x+2) = 0

=> (x+2)(x+3) = 0

=> x+2 = 0 or x+3 = 0

=> x = -2 or -3

Zeroes are -2 and -3

Their ratio = -3:-2 = -3/-2 = 3/2 = 3:2

Verified the given relations in the given problem

Used formulae:-

  • The standard quadratic polynomial is ax²+bx+c

  • Sum of the zeroes = -b/a

  • Product of the zeroes = c/a

  • To get the zeores of a polynomial we equate it to zero.
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