Question

Find the zeroes of the polynomial f(x) = 4x² + 8x and verify the relationship between the zeroes and its coefficients.

Plsssssssssssssssss help me ______ :)

Answer 1

Answer:

Find the zeroes of the quadratic polynomial 4x^2+8x and verify the relationship between the zeroes and the coefficients

4x² + 8x  

= 4x ( x + 2)

To find Zeroes

4x ( x + 2) = 0

=> x = 0 & x = -2

Sum of Zeroes = 0 + (-2) = -2

Product of zeroes = 0 * (-2) = 0

Comparing 4x² + 8x

with ax² + bx + c

a = 4   , b = 8   , c = 0

Sum of Zeroes = -b/1 = - 8/4 = -2

Product of Zeroes = c/a = 0/4 = 0

Sum & product of zeroes relationship is verified as results matches.

Step-by-step explanation:

Answer 2

S O L U T I O N :

We have quadratic polynomial f(x) = 4x² + 8x & zero of the polynomial f(x) = 0.

$\underline{\underline{\tt{Using\:by\:factorization\:\:method\::}}}$

→ 4x² + 8x = 0

→ 4(x² + 2x) = 0

→ 4x(x + 2) = 0

→ x + 2 = 0/4x

→ x + 2 = 0

→ x = -2   Or   x = 0

∴ α = -2 & β = 0 are two zeroes of the given polynomial .

As we know that given quadratic polynomial compared with ax² + bx + c;

• a = 4
• b = 8
• c = 0

Now,

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha + \beta = \dfrac{-b}{a} = \bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\mapsto\tt{-2 + 0 = \dfrac{-8}{4}}$

$\mapsto\tt{-2 + 0 = \cancel{\dfrac{-8}{4}}}$

$\mapsto\bf{-2 = -2}$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha \times \beta = \dfrac{c}{a} = \bigg\lgroup \dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\mapsto\tt{-2 \times 0 = \dfrac{0}{4}}$

$\mapsto\bf{0 = 0}$

Thus,

The relationship between zeroes & coefficient are verified .