NCERT · Class 10 · Mathematics · PolynomialsFind the zeros of the quadratic polynomial $4u^2 + 8u$, and verify the relationship between the zeros and the coefficients.

To find the zeros of $p(u) = 4u^2 + 8u$, we factorize the expression:

1. Factorization: $4u^2 + 8u = 0$ Taking $4u$ as common: $4u(u + 2) = 0$

   So, $4u = 0$ or $u + 2 = 0$ $u = 0$ or $u = -2$ Therefore, the zeros are $\alpha = 0$ and $\beta = -2$.

2. Verification: Comparing $4u^2 + 8u$ with the general form $au^2 + bu + c$, we get: $a = 4, b = 8, c = 0$.

   • Sum of zeros: $\alpha + \beta = 0 + (-2) = -2$ Formula: $-\frac{b}{a} = -\frac{8}{4} = -2$ Hence, Sum of zeros $= -\frac{b}{a}$ (Verified).

   • Product of zeros: $\alpha \cdot \beta = 0 \times (-2) = 0$ Formula: $\frac{c}{a} = \frac{0}{4} = 0$ Hence, Product of zeros $= \frac{c}{a}$ (Verified).