Question

find the zeros of the quadratic polynomial 4 x square - 4 x minus 3 and verify the rational numbers rational between the zeros and coefficients

Answer 1

4x^2-4x-3=0
4x^2-6x+2x-3=0
taking common
2x(2x-3)+1(2x-3)=0
(2x+1)(2x-3)=0
x = -1/2,3/2

now verifying the relationship
from equation
aplha+beeta=-b/a
-(-4)/4=1
from zeros
-1/2+3/2=2/2=1

from equation
alpha× beeta=c/a
= -3/4

from zeros
-1/2× 3/2= -3/4


hope it helps you.

Answer 2

Hey there !!

Answer:

→ -1/2 and 3/2 .

Step-by-step explanation:

Let the given polynomial be denoted by f(x) . Then,

f(x) = 4x² - 4x - 3 .

= 4x² - 6x + 2x - 3 .

= 2x( 2x - 3 ) + 1( 2x - 3 ) .

= ( 2x + 1 ) ( 2x - 3 ) .

∴ f(x) = 0 .

⇒ ( 2x + 1 ) ( 2x - 3 ) = 0 .

⇒ 2x + 1 = 0 .      or      2x - 3 = 0 .

⇒ x = $\bf \pink{ \frac{-1}{2} }$ .   or   x = $\bf \orange{ \frac{3}{2} }$ .

So, the zeroes of f(x) are -1/2 and 3/2 .

Sum of zeros = $\bf{ \frac{-1}{2} + \frac{3}{2} = \frac{2}{2}= 1}$

$\implies 1 = \frac{ coefficient \: of \: x }{ coefficient \: of \: {x}^{2} }$

Product of zeroes = $\bf \frac{-1}{2} \times \frac{3}{2} = \frac{-3}{4} .$

$\implies \frac{-3}{4} = \frac{ Constant \: term }{ Coefficient \: of \: {x}^{2} }$

Hence, it is solved .

THANKS

#BeBrainly .