Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials: 2x^2

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:

2x^{2}+\frac{7}{2}x+\frac{3}{4}

Answers (1)

Answer.  \left [\frac{-3}{2},\frac{-1}{4} \right ]

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

2x^{2}+\frac{7}{2}x+\frac{3}{4}

Multiply by 4

8x^2 + 14x + 3 =0

8x^2 + 12x + 2x + 3 =0

4x(2x + 3) + 1(2x + 3)=0

(2x + 3) (4x + 1) =0

2x + 3 = 0                      4x + 1 = 0

x=\frac{-3}{2}                              x=\frac{-1}{4}                

Hence, \left [\frac{-3}{2},\frac{-1}{4} \right ]  are the zeroes of the polynomial

Here, a = 2, b =\frac{7}{2} , c = \frac{3}{4}

Sum of zeroes \frac{-b}{a}=\frac{-7}{2 \times 2}=\frac{-7}{4}

Here, =\frac{-3}{2}-\frac{1}{4}=\frac{-6-1}{4}=\frac{-7}{4}=\frac{-b}{a}

Product of zeroes =\frac{c}{a}=\frac{3}{4 \times 2}=\frac{3}{8}

Here, -\frac{3}{2}\times \frac{-1}{4}=\frac{3}{8}=\frac{c}{a}

Posted by

infoexpert21

View full answer

Similar Questions

Latest Question