spriya
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Find the zeroes of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.
(iv) 4u2 + 8u

Answer :

Mathexpert
Let 4u² + 8u = 0

4u(u + 2) = 0

4u = 0 or u + 2 = 0
u = 0 or u = -2

sum of the roots = -b/a  = -8/4 = -2
But sum of the roots = 0 + (-2) = -2

Product of the roots = c/a = 0 (As there is no c in the equation)
But, product of the roots = 0 × –2 = 0

Given:

We have given an equation

To Find:

We have to find the zeroes of quadratic equation?

Step-by-step explanation:

  • We have the given quadratic equation which is

       [tex]4u^2+8u[/tex]

  • For finding the zeroes of quadratic equation we will simplify the equation by factorization

       take the common terms out from above equation

       [tex]4u(u+8)[/tex]

  • Now equate it with 0 we get

      [tex]4u=0,u+2=0\\u=0,u=-2[/tex]

  • Hence we get the zeroes are 0,-2
  • Sum of zeroes is 0+(-2)=-2
  • Product of zeroes=0(-2)=0

      Now we will verify the relation by using formula

  • comparing the given quadratic equation with [tex]ax^2+bx+c=0[/tex]

      We get a=4 , b = 8

  • Now sum of zeroes is given by the formula
  • [tex]Sum=\frac{-b}{a} =\frac{8}{4} =-2[/tex]
  • Product of zeroes is given by the foumla

      [tex]\textrm{Product of zeroes}=\frac{c}{a} =0[/tex]

Hence, the zeroes are 0,-2 and sum is -2,product is 0

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