Question

Let ( x(square)+ 3x +1) over (2 x(square) -5x+2) = k ----------be y

1) Express y in the form of a x(square) + bx +c =0

2) If -3/4 is a root of y, find the value of k


I don't really understand why the answer is like that. Can you explain?

(x^2+3x+1)/(2x^2-5x+2) = k
x^2+3x+1 = 2kx^2 - 10kx+2k
(1-2k)x^2 + (3+10k)x + (1-2k) = 0

since -3/4 is a root,

(1-2k)(9/16) + (3+10k)(-3/4) + (1-2k) = 0
-1/16 (170k+11) = 0
k = -11/170

Answer 1

( x ^ 2 + 3 x + 1 ) / ( 2 x ^ 2 - 5 x + 2 ) = k Multiply both sides by 2 x ^ 2 - 5 x + 2

x ^ 2 + 3 x + 1 = k ( 2 x ^ 2 - 5 x + 2 )

x ^ 2 + 3 x + 1 = 2 k x ^ 2 - 5 k x + 2 k

x ^ 2 + 3 x + 1 - 2 k x ^ 2 + 5 k x - 2 k = 0

( 1 - 2 k ) x ^ 2 + ( 3 + 5 k ) x + 1 - 2 k = 0

y = ( 1 - 2 k ) x ^ 2 + ( 3 + 5 k ) x + 1 - 2 k = 0


Now we try to find k

We now :

for x = - 3 / 4 , y = 0


0 = ( 1 - 2 k ) x ^ 2 + ( 3 + 5 k ) x + 1 - 2 k

( 1 - 2 k ) x ^ 2 + ( 3 + 5 k ) x + 1 = 0

( 1 - 2 k ) * ( - 3 / 4 ) ^ 2 + ( 3 + 5 k ) * ( - 3 / 4 ) + 1 - 2 k = 0

( 1 - 2 k ) * 9 / 16 + ( 3 + 5 k ) * ( - 3 / 4 ) + 1 - 2 k = 0 Multiply both sides by 16

( 1 - 2 k ) * 9 * 16 / 16 + ( 3 + 5 k ) * ( - 3 / 4 ) * 16 + 1 * 16 - 2 k * 16 = 0

( 1 - 2 k ) * 9 + ( 3 + 5 k ) * ( - 3 / 4 ) * 4 * 4 + 16 - 32 k = 0

( 1 - 2 k ) * 9 + ( 3 + 5 k ) * ( - 3 ) * 4 + 16 - 32 k = 0

( 1 - 2 k ) * 9 + ( 3 + 5 k ) * ( - 12 ) + 16 - 32 k = 0

1 * 9 - 2 k * 9 + 3 * ( - 12 ) + 5 k * ( - 12 ) + 16 - 32 k = 0

9 - 18 k - 36 - 60 k + 16 - 32 k = 0

- 110 k - 11 = 0

- 110 k = 11 Divide both sides by - 100

k = 11 / - 110

k = - 1 / 10

Answer 2

Correction:


- 110 k = 11 Divide both sides by - 110

k = 11 / - 110

k = - 1 / 10