Question

find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients
4u²+8u​

Answer 1

AnsWer :

u = 0 and u= -2.

Given :

• We have equation, 4u²+8u.

Concepts Required :

$\blacksquare \tt \: Sum \: of \: zeroes. \\ \tt\alpha + \beta = \frac{ - b}{a} = \frac{Coefficient \: of \: x}{Coefficient \: of \: {x}^{2} }$

$\blacksquare \tt \: Product \: of \: zeroes. \\ \tt \alpha . \beta = \frac{c}{a} = \frac{Constant \: term}{Coefficient \: of \: {x}^{2} }$

General Equation, Of Quadratic Polynomial.

$\tt a {x}^{2} + bx + c = 0. \: \: \red{where \: a \neq \: 0.}$

Solution :

We have equation,

$\tt4 {u}^{2} + 8u = 0.$

Let's Compare with General Equation,

$\tt a {x}^{2} + bx + c = 0.$

Where,

• a = 4.
• b = 8.
• c = 0.

$\implies \tt4 {u}^{2} + 8u = 0.$

$\implies \tt4 u(u + 2) = 0.$

$\implies \tt(u + 2)(4u) = 0.$

Either,

$\implies \tt4 u= 0.$

$\implies \tt u = 0.$

Or,

$\implies \tt u + 2= 0.$

$\implies \tt u= - 2.$

$\rule{200}3$

Let's Verify,

Let,

$\bullet \: \tt \alpha = 0. \: \: \: \: \: \: \: \: \red{and} \: \: \: \: \: \: \: \: \bullet \: \beta = - 2.$

So,

Sum of the Zeros.

$\tt \implies \alpha + \beta = \frac{ - b}{a}$

$\tt \implies - 2 + 0 = \frac{ - 8}{4}$

$\tt \implies - 2 = - 2.$

Product of the Zeros.

$\tt \implies \alpha. \beta = \frac{c}{a}$

$\tt \implies - 2 \times 0 = \frac{0}{4}$

$\tt \implies 0 = 0.$

Therefore, the value of u = 0 or, u = -2.

Answer 2

$\large \underline{ \blue{ \boxed{ \bf \green{Required \: Answer}}}}$