SOLUTION: I've been working on an assignment and I'm stumped. I have to solve quadratic equations with the following steps: a) Move the constant term to the right side of the equation. b)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I've been working on an assignment and I'm stumped. I have to solve quadratic equations with the following steps: a) Move the constant term to the right side of the equation. b)      Log On


   

Question 157876: I've been working on an assignment and I'm stumped. I have to solve quadratic equations with the following steps:
a) Move the constant term to the right side of the equation.
b) Multiply each term in the equation by four times the coefficient of the x2 term.
c) Square the coefficient of the original x term and add it to both sides of the equation.
d) Take the square root of both sides.
3) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.
f) Set the left side of the equation equal to the negative square root of thenumber on the right side of the euqation and solve for x.
I have two equations I'm stuck on:
1) x2 - 2x - 13 = 0
2) 2x2 - 3x - 5 = 0
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
a) Move the constant term to the right side of the equation.
1)+x%5E2+-+2x+-+13+=+0
x%5E2+-+2x+=+13
b) Multiply each term in the equation by four times the coefficient of the x%5E2 term.
the coefficient of the x%5E2 is 1 and four times the coefficient will be 4%2A1=4, so you will have:
x%5E2%2A4+-+2x%2A4+=+13%2A4
4x%5E2+-+8x+=+52
c) Square the coefficient of the original x+term and add it to both sides of the equation.
the coefficient of the original x+term is -2
4x%5E2+-+8x+%2B+%28-2%29%5E2++=+52+%2B+%28-2%29%5E2
4x%5E2+-+8x+%2B+4+=+52+%2B+4…then you have
4x%5E2+-+8x+%2B+4+=+56
%282x-2%29%5E2=+56
d) Take the square root of both sides.
sqrt%28%282x-2%29%5E2%29 = +-sqrt%2856%29
2x-2 = +-%287.48%29

e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.
2x-2+=+7.48
2x+=+7.48%2B2
2x+=+9.48
x+=+9.48%2F2
x+=+4.74

f) Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x.
2x-2+=+-7.48
2x+=+-7.48%2B2
2x+=+-5.48
x+=+-5.48%2F2
x+=+-2.74


2) 2x%5E2+-+3x+-+5+=+0
a) Move the constant term to the right side of the equation.
2x%5E2+-+3x+=+5+
b) Multiply each term in the equation by four times the coefficient of the x%5E2 term.
the coefficient of the x%5E2 is 2 and four times the coefficient will be 4%2A2=+8, so you will have:
2x%5E2%2A8+-+3x%2A8+=+5%2A8+
16x%5E2+-24%2Ax+=+40+

c) Square the coefficient of the original x+term and add it to both sides of the equation.
the coefficient of the original x+term is -3
16x%5E2+-+24x+%2B+%28-3%29%5E2+=+40+%2B++%28-3%29%5E2+
16x%5E2+-+24x+%2B+9+=+40+%2B++9+
16x%5E2+-+24x+%2B+9+=+49+
%284x-3%29%5E2=+49

d) Take the square root of both sides
sqrt%28%284x-3%29%5E2%29=+sqrt%2849%29
4x-3 = +-%287%29

e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.
4x-3 = +%287%29
4x+=+7%2B3+
4x+=+10+
x+=+10%2F4
x+=+2.5

f) Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x.
4x-3 = -%287%29
4x+=+-7%2B3+
4x+=+-4+
x+=+-4%2F4
x+=+-1