Question

Find the zeros of quadratic polynomial 8x^2-22x-21.

Answer 1

Hey!

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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!

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Hope it helps...!!

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Answer 2

hope this helps you ☺☺