Question 1 (iv)
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

$4 u^{2} + 8 u$

Open in App

Solution

Comparing given polynomial with the general form $a x^{2} + b x + c$ ,

We get, a = 4, b = 8 and c = 0
We have, $4 u^{2} + 8 u$ = 4u(u + 2)
The value of $4 u^{2} + 8 u$ is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2
Therefore, the zeroes of $4 u^{2} + 8 u$ are 0 and - 2.
Sum of zeroes = 0 + (-2) = -2 = $\frac{- (8)}{4} = \frac{- (C o e f f i c i e n t o f u)}{C o e f f i c i e n t o f u^{2}}$
Product of zeroes = 0 $\times$ (-2) = 0 = $\frac{0}{4} = \frac{C o n s t a n t t e r m}{C o e f f i c i e n t o f u^{2}}$

Suggest Corrections

Similar questions

4 u^{2} + 8 u

4 u^{2} + 8 u

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^{2} - 2 x - 8$

(ii) $4 s^{2} - 4 s + 1$

(iii) $6 x^{2} - 3 - 7 x$
(iv) $4 u^{2} + 8 u$

(v) $t^{2} - 15$

(vi) $3 x^{2} - x - 4$

2 y cos x = y^{2} + 2 c

4 x^{2} - 4 x + 1

6 x^{2} - 3 - 7 x

4 u^{2} + 8 u

t^{2} - 15

3 x^{2} - x - 4

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program