Math, asked by nityarabadiya3, 5 hours ago

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and their coefficients. (i) 63 - 2x - x2 (ii) 49x2 - 81​

Answers

Answered by tennetiraj86
2

Step-by-step explanation:

Given :-

(i) 63 - 2x - x²

(ii) 49x² - 81

To find :-

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and their coefficients.

Solution :-

i)

Given Quadratic Polynomial is 63-2x-x²

Let P(x) = 63-2x-x²

On writting this in the standard form

P(x) = -x²-2x+63

Finding the zeroes :-

To get the zeroes we write P(x) = 0

=> -x²-2x+63 = 0

=> -(x²+2x-63) = 0

=> x²+2x-63 = 0

=> x²+9x-7x-63 = 0

=> x(x+9)-7(x+9) = 0

=> (x+9)(x-7) = 0

=> x+9 = 0 or x-7 = 0

=> x = -9 or x = 7

The zeroes are -9 and 7

Finding the relationship between the zeroes and the coefficients:-

P(x) = -x²-2x+63

On Comparing this with the standard quadratic Polynomial ax²+bx+c

a = -1

b= -2

c = 63

The zeroes are -9 and 7

Let α = -9 and β = 7

Sum of the zeroes = α +β

=> -9+7

=> -2

=> α + β = -2

Sum of the zeroes = -b/a

=> -(-2)/-1

=> 2/-1

=> -2

Sum of the zeroes =α+ β = -b/a

Product of the zeroes = α β

=> α β = (-9)(7)

=> α β = -63

Product of the zeroes = c/a

=> 63/-1

=> -63

Product of the zeroes = α β = c/a

________________________________

ii)

Given Quadratic Polynomial is 49x²-81

Let P(x) = 49x²-81

Finding the zeroes :-

To get the zeroes we write P(x) = 0

=> 49x²-81= 0

=>7²x²-9²= 0

=>(7x)²-9²= 0

=> (7x+9)(7x-9) = 0

Since (a+b)(a-b) = a²-b²

Where , a = 7x and b = 9

=> 7x+9 = 0 or 7x-9 = 0

=> 7x = -9 or 7x= 9

=> x = -9/7 or x = 9/7

The zeroes are -9/7 and 9/7

Finding the relationship between the zeroes and the coefficients:-

P(x) =49x²-81

On Comparing this with the standard quadratic Polynomial ax²+bx+c

a = 49

b= 0

since there is no term with coefficient of x

c =-81

The zeroes are -9/7 and 9/7

Let α = -9/7 and β = 9/7

Sum of the zeroes = α +β

=> (-9/7)+(9/7)

=> (-9+9)/7

=> 0/7

=> 0

=> α + β = 0

Sum of the zeroes = -b/a

=> 0/49

=>0

Sum of the zeroes =α+ β = -b/a

Product of the zeroes = α β

=> α β = (-9/7)(9/7)

=> α β = (-9×9)/(7×7)

=>α β=-81/49

Product of the zeroes = c/a

=> -81/49

Product of the zeroes = α β = c/a

Answer:-

i) The zeroes are -9 and 7

ii)The zeroes are -9/7 and 9/7

Verified the given relations in the given problems.

Used formulae:-

  • The standard quadratic polynomial is ax²+bx+c

  • Sum of the zeroes =α+ β = -b/a

  • Product of the zeroes = α β = c/a

  • (a+b)(a-b) = a²-b²

  • To get the zeroes of P(x) we can write it as P(x) = 0.
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