Algebra

posted by John

I got a couple I want to check my work :
Both are Solve using the quadratic formula :
1) x^2-3x=7x-2
x^2-3x-7x+2=7x-7x-2+2
x^2-10x+2=0
x = 10 +- sqrt10^2 - 4(1)(2) / 2(1)
x = 10 +- sqrt100 - 8 / 2
x = 10 +- sqrt92 / 2
x = 10 +- 2sqrt23 /2
x = 5 +- sqrt23
x = 5 + sqrt23 and x = 5 - sqrt23

2) x^2-7x-1= -7
x^2-7x+7-1 = -7+7
x^2-7x+6 = 0
x = 7 +- sqrt7^2 - 4(1)(6) / 2(1)
x = 7 +- sqrt49-24 / 2
x = 7 +- sqrt25 / 2
x = 7 +- 5 / 2
x = 7+5 / 2 , x = 12 / 2 , x = 6
x = 7-5 / 2 , x = 2 / 2 , x = 1

1. DrBob222

The second one looks ok. I think step 7 of problem 1 has an error.

2. John

Ok so what I did was
x = 10 +- 2sqrt23 / 2
I divided 10 by 2 and 2 by 2
Thats how i got
x = 5 +- sqrt23

3. DrBob222

1) x^2-3x=7x-2
x^2-3x-7x+2=7x-7x-2+2
x^2-10x+2=0
x = 10 +- sqrt10^2 - 4(1)(2) / 2(1)
x = 10 +- sqrt100 - 8 / 2
x = 10 +- sqrt92 / 2 OK to here. See below.

x = 10 +- 2sqrt23 /2
x = 5 +- sqrt23
x = 5 + sqrt23 and x = 5 - sqrt23

x = (10+-sqrt 92)/2
x = (10+-sqrt 2(46)/2
(What you are doing here is dividing 10 by 2, which is correct, as well as 2/2, which also is correct. What is NOT correct is dividing 46 by 2 (you are using the divisor of 2 twice but the divisor of 2 disappeared when you used it to divide the 10 and 2 of 2*46).
x = (5+-sqrt 46)

4. tamesha

7x-2=22.5

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