Question

18. Find the zeroes of the quadratic polynomial 4x2 – 6 – 8x and verify the relationsp between the zeroes and the coefficients of the polynomial.

Answer 1

Answer:

x = (2 ± √10) / 2

Step-by-step explanation:

Quadratic : 4x² - 8x - 6 = 0, simplifying

We get, 2x² - 4x -3 =0 → x² - 2x - 3/2 =0

Using x = -b ± √b²-4ac / 2a to find x,

Solving, we get, x = 2 ± √10 / 2

From the given quadratic we get,

→ Sum of zeros of quadratic = 2 and Product of zeros = -3/2

VERIFYING,

→ Sum = (2 + √10 )/ 2  +  (2 - √10) / 2 = 4/2 = 2

→ Product = (2 + √10 )/ 2  x   (2 - √10) / 2 = (4 -10/4) = -6/4 = -3/2

Answer 2

Answer:

1 +  √10/2

1 - √10/2

Step-by-step explanation:

Find the zeroes of the quadratic polynomial 4x2 – 6 – 8x and verify the relationsp between the zeroes and the coefficients of the polynomial.

4x² - 6 - 8x

= 4x² - 8x - 6

x =  (8 ± √8² - 4*4(-6) )/(2*4)

x = (8 ± 4√10)/8

x = 1 ± √10/2

1 +  √10/2

1 - √10/2

Sum of roots = 1 + √10/2  + 1 - √10/2  = 2

Sum of roots =  - (coefficient of x)/(coefficient of x²) =  - (-8)/4 = 2

Products of roots = (1 + √10/2)(1 - √10/2) = 1 - 10/4 = -6/4

Products of roots = constant/(coefficient of x²) = -6/4