Question

Find the zeroes of the quadratic polynomial 8x² + 3x - 5 and verify the relationship between the zeroes and the coefficients.​

Answer 1

Hi,

Explanation:

`8x^2+3x-5=8x^2+8x-5x-5\\\\=8x(x+1)-5(x+1)\\\\=(x+1)(8x-5)\\\\Zeroes\ are\ x=-1\ and\ x=(5)/(8) \\\\x_1+x_2=-1+(5)/(8) =-(3)/(8) \\\\x_1 * x_2=- (5)/(8) \\\\a=8, b= 3, c=-5\\\\-(b)/(a) =(3)/(8) =x_1+x_2\\\\(c)/(a) =(-5)/(8) =x_1*x_2\\\\`

Answer 2

see explanation

Explanation:

to find the zeros let the function equal zero , that is

8x² + 3x - 5 = 0

consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 8 × - 5 = - 40 and sum = + 3

the factors are + 8 and - 5

use these factors to split the x- term

8x² + 8x - 5x - 5 = 0 ( factor the first/second and third/fourth terms )

8x(x + 1) - 5(x + 1) = 0 ← factor out (x + 1) from each term

(x + 1)(8x - 5) = 0

equate each factor to zero and solve for x

x + 1 = 0 ( subtract 1 from both sides )

x = - 1

8x - 5 = 0 ( add 5 to both sides )

8x = 5 ( divide both sides by 8 )

x =
`(5)/(8)`

zeros are x = - 1 and x =
`(5)/(8)`

for a quadratic equation in standard form

ax² + bx + c = 0 ( a ≠ 0 )

then the relationship between the zeros and the coefficients is

sum of zeros =
`(-b)/(a)`

product of zeros =
`(c)/(a)`

here a = 8 , b = 3 , c = - 5

sum =
`(-3)/(8)` = -
`(3)/(8)` and product =
`(-5)/(8)` = -
`(5)/(8)`

check with actual zeros

sum = - 1 +
`(5)/(8)` = -
`(8)/(8)` +
`(5)/(8)` = -
`(3)/(8)` ← agrees with above

product = - 1 ×
`(5)/(8)` = -
`(5)/(8)` ← agrees with above

showing the relationship between zeros and coefficients is correct