Question

Class - 10
Chapter - Polynomials
Q. 16

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Answer 1

Answer:

Question 1.

If the sum of zeroes of the quadratic polynomial 3x2 – kx + 6 is 3, then find the value of k. (2012)

Solution:

Here a = 3, b = -k, c = 6

Sum of the zeroes, (α + β) = −ba = 3 …..(given)

⇒ −(−k)3 = 3

⇒ k = 9

Question 2.

If α and β are the zeroes of the polynomial ax2 + bx + c, find the value of α2 + β2. (2013)

Solution:

Important Questions for Class 10 Maths Chapter 2 Polynomials 1

Answer 2

Step-by-step explanation:

Given :-

A quadratic polynomial f(x) = x²-3x-2

α and β are the zeroes of f(x).

To find :-

Find the quardratic polynomial whose zeores are

1/(2α+ β) and 1/(2β+α)?

Solution :-

Given quadratic polynomial f(x) = x²-3x-2

=> f(x) = x²-x-2x-2

=> f(x) = x(x-1)-2(x-1)

=> f(x) = (x-1)(x-2)

To find the zeores of f(x) we write f(x) = 0

=> f(x) = (x-1)(x-2) = 0

=> (x-1) = 0 or (x-2) = 0

=> x = 1 or x=2

So the Zeroes of f(x) are 1 and 2

Let α = 1 and β = 2

Now ,

The value of 1/(2α+ β)

= 1/(2(1)+2)

= 1/(2+2)

= 1/4

and the value of 1/(2β+α)

= 1/(2(2)+1)

= 1/(4+1)

= 1/5

The zeores are 1/4 and 1/5

now,

We know that

The quardratic polynomial whose zeores are α and β is K[x²-(α + β)x+α β]

The Quadratic Polynomial whose zeores are

1 /(2α+ β) and 1/(2β+α)

= K[x²-{(1/4)+(1/5)}x+(1/4)(1/5)]

= K[x²-{(5+4)/20}x+(1/20)]

= K[x²-(9/20)x+(1/20)]

=K[(20x²-9x+1)/20]

If K = 20 then

The required Polynomial is 20x²-9x+1

Answer:-

The required quadratic polynomial whose zeores

are 1 /(2α+ β) and 1/(2β+α) is 20x²-9x+1

Used formulae:-

• The standard quadratic Polynomial is ax²+bx+c

• The quardratic polynomial whose zeores are α and β is K[x²-(α + β)x+α β]

• To get Zeroes we write given Polynomial equal to zero.

• Sum of the zeroes = -b/a

• Product of the zeroes = c/a