maths
posted by Jayden Haddy on .
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^25x+2) = k
x^2+3x+1 = 2kx^2  10kx+2k
(12k)x^2 + (3+10k)x + (12k) = 0
since 3/4 is a root,
(12k)(9/16) + (3+10k)(3/4) + (12k) = 0
1/16 (170k+11) = 0
k = 11/170

( 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 
Correction:
 110 k = 11 Divide both sides by  110
k = 11 /  110
k =  1 / 10