Question

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

$4u2 + 8u$
​

Answer 1

Step-by-step explanation:

$4 {u}^{2} + 8u = 0 \\ \\ 4u(u + 2) = 0 \\ 4u = 0 \: \: \: u + 2 = 0 \\ u = 0 \: \: \: \: u = - 2$

$\alpha + \beta = \frac{ - b}{a} \\ 0 + ( - 2) = \frac{ - 8}{4} \\ - 2 = - 2 \\ \\ \alpha \beta = \frac{c}{a} \\ 0 \times - 2 = \frac{0}{ 4} \\ 0 = 0$

HOPE YOU GET YOUR ANSWER

Answer 2

$\underline{\underline{\sf SOLUTION :}}$

We are given the quadratic polynomial as 4u² + 8u = 0 and we need to find the zeros of the given quadratic polynomial and verify the relationship between the zeros and the coefficients. So, first compare the given quadratic polynomial with general equation ax² + bx + c = 0 where a = 4, b = 8 and c = 0 then take 4 common from the given quadratic polynomial and hence we can obtain our zeros. So,let's do it :

$\qquad\dag\:\: \underline{\textsf {Zeros of the given quadratic polynomial : }}$

$:\implies \sf 4u^2 + 8u = 0$

$:\implies \sf 4u(u + 2) = 0$

$:\implies \sf (4u)(u + 2) = 0$

$:\implies \underline{ \boxed{\sf u= 0 \: or\: - 2}}$

Hence, we have find the zeros of the given quadratic polynomial and now we have to verify the relationship between the zeros and the coefficients. Now,first assume α = 0 and β = -2. So, let's do it :

$\qquad\dag\:\: \underline{\textsf {Relationship between the zeros and the coefficients : }}$

$\star\: \underline{\textbf {Sum of zeros :}}$

$\dashrightarrow\:\:\sf \alpha + \beta = \dfrac{-b}{a}$

$\dashrightarrow\:\:\sf 0 + ( - 2) = \dfrac{-8}{4}$

$\dashrightarrow\:\: \underline{ \boxed{\sf - 2 = - 2 }}$

$\star\: \underline{\textbf {Product of zeros :}}$

$\dashrightarrow\:\:\sf \alpha \times \beta = \dfrac{c}{a}$

$\dashrightarrow\:\:\sf 0 \times (- 2) = \dfrac{0}{4}$

$\dashrightarrow\:\: \underline{ \boxed{\sf 0 = 0}}$

$\qquad \quad\dag \: \textsf{\textbf{Hence, verified !}} \: \dag$