Question

One of the roots of equation x^(2)-(3p-4)x+11p-5=0 is 8. Find the other root. Find the value of p.

Answer 1

p = 7

Explanation:

x²-(3p-4)x+11p-5=0 ...(1) Root: 8, x-8 is factor

(x-8) (x- (11p-5)/8) = 0

x² -8x - ((11p-5)/8)*x + (11p-5) = 0

x² - (8 + (11p-5)/8) x + 11p-5 = 0 ----(2)

compare(1) and (2): 3p-4 = 8 + (11p-5)/8

24p - 32 = 64 + 11p - 5

13p = 91

p = 7

check: x²-(3p-4)x+11p-5 = x²- 17x + 72 = (x-8)(x-9)

Answer 2

other root: x = 9

p = 7

Explanation:

first, find p:

(8)² - (3p - 4)8 + 11p -5 = 0

64 - (24p - 32) + 11p - 5 = 0 –––––> 64 - 24p + 32 + 11p - 5 = 0

91 - 13p = 0

-13p = -91

p = 7

plug in p to find the polynomial:

x² - (3(7) - 4)x + 11(7) - 5 = 0

x² - (21 - 4)x + 77 - 5 = 0

x² - 17x + 72

find the roots:

we know that 8 is one of the roots, so one factor is (x - 8)

the other number needs to be something that equals -17 when added with -8 and 72 when multiplied by -8

that number is -9

so the poly nomial factors to be: (x-8)(x-9)

so the roots are x = 8 and x = 9