Question

find the zeros of the quadratic polynomial 8 x square - 21 - 22 x and verify the relationship between zeroes and coefficients of the polynomials

Answer 1

8x^2-22x-21=0
8x^2-28x+6x-21=0
4x(2x-7)+3(2x-7)=0
(4x+3)(2x-7)=0
either            or
4x+3=0         2x-7=0
4x=-3            2x=7
x=-3/4             x=7/2
here a=8   b=-22   c=-21  α=-3/4 β=7/2
α+β=-b/a                                   αβ=c/a
-3/4+7/2=-(-22/8)                      -3/4*7/2=-21/8
-3/4+14/4=(22/8)                       -21/8=-21/8
11/4=11/4
hence verified

Answer 2

Hey!

_______________

8x^2 - 21 - 22x

8x^2 - 22x - 21

8x^2 - 28x + 6x - 22x

(8x^2 - 28x) + (6x - 22x)

4x ( 2x - 7) + 3 (2x - 7)

(4x + 3) (2x - 7)

Now,

4x + 3 = 0
x = -3/4

2x - 7 = 0
x = 7/2

Comparing 8x^2 - 22x - 21 with ax^2 + bx + c

a = 8 , b = -22 , c = -21

Alpha ( @ ) = -3/4
Beta ( ß ) = 7/2

Sum of zeroes ( @ + ß ) = - b/a

-3/4 + 7/2 = - ( - 22) / 8

-3 + 14 /4= 22/8

11/4 = 11/4


Product of zeroes ( @ × ß ) = c / a

- 3/4 × 7/2 = -21 / 8

-21 / 8 = -21/8



Hence, Verified!

_______________

Hope it helps...!!!