CAN SOMEONE PLEASE CHECK TO SEE IF I HAVE DONE THESE CORRECTLY. Thanks for your help.
Solve by completing the square
1. X^2-6x=16
X^2-6x + (-6/2) ^2=16+ (-6/2) ^2
X^2-6x+36/4= 16+36/4
(X-6/2)=100/4
X-6/2= +or - 10/2
X=6 + or - 10/2
X=-4, X=8
2. 2x^2-3x+1
2x^2-3x=-1
X^2-3/2x=-1/2
X^2-3/2x +(-3/2)^2 =-1/2+(-3/2)^2
X^2-3/2+9/4 =-1/2+9/4
X+3/2=7/4
X+3/2=+ or - sqrt7/4
X=-3 + or – sqrt 7/2
X= -3+sqrt7/2 , x=-3-sqrt7/2
Put the values in the letters and see if you get the right answer
For example
Put the value for your first answer as
-4 squared - 6(8) = 16
See if it works in this way
No1 in line 4
why don't your reduce 6/2 to 3
line 4 would read (x-3)^2 = 25
then x-3 = +/- √25
x-3 = +/- 5
then x=8 or x = -2
no 2 you have to take 1/2 of 3/2 which would be 3/4.
then 3/4 squared is 9/16
so line 4 should read
x^2 - (3/2)x + 9/16 = -1/2 + 9/16
.
.
.
(x-3/4)^2 = 1/16
..
x=1 or x=1/2
To check the solutions of the equations, let's substitute the values back into the original equations and see if they hold true.
1. For the equation x^2 - 6x = 16:
Let's substitute x = -4:
(-4)^2 - 6(-4) = 16
16 + 24 = 16
40 = 16 (This is not true, so -4 is not a correct solution.)
Let's substitute x = 8:
(8)^2 - 6(8) = 16
64 - 48 = 16
16 = 16 (This is true, so x = 8 is correct.)
2. For the equation 2x^2 - 3x + 1 = 0:
Let's substitute x = -3 + sqrt(7)/2:
2(-3 + sqrt(7)/2)^2 - 3(-3 + sqrt(7)/2) + 1 = 0
Simplifying further, we find:
1 - 3sqrt(7)/2 + 7/2 + 9/2 - 3sqrt(7)/2 + 1 = 0
2/2 + 9/2 = 0
11/2 = 0 (This is not true, so -3 + sqrt(7)/2 is not a correct solution.)
Let's substitute x = -3 - sqrt(7)/2:
2(-3 - sqrt(7)/2)^2 - 3(-3 - sqrt(7)/2) + 1 = 0
Simplifying further, we find:
1 + 3sqrt(7)/2 + 7/2 - 9/2 + 3sqrt(7)/2 + 1 = 0
2/2 - 9/2 = 0
-7/2 = 0 (This is not true, so -3 - sqrt(7)/2 is not a correct solution.)
Therefore, the solutions you provided for both equations seem to be incorrect. Please recheck your calculations and revise the solutions accordingly.