SOLUTION: I am stuck on the following quadratic equation:
2x^2-3x-1=0
I move the 1 over like this:
2x^2-3x=1
and that's where I get stuck: how do you create a third term to the l
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: I am stuck on the following quadratic equation:
2x^2-3x-1=0
I move the 1 over like this:
2x^2-3x=1
and that's where I get stuck: how do you create a third term to the l
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Question 33311This question is from textbook Practical Algebra: A Self-Teaching Guide
: I am stuck on the following quadratic equation:
2x^2-3x-1=0
I move the 1 over like this:
2x^2-3x=1
and that's where I get stuck: how do you create a third term to the left side of the equation like that will later generate -3x? Do you take square half the value of the second term? (-3x / 2)^2, which results in 9/4x?
Thanks so much for your help. This question is from textbook Practical Algebra: A Self-Teaching Guide
You can put this solution on YOUR website! Solving by completing the square:
Divide by 2 to get:
x^2-(3/2)x =1/2
Complete the square on the left side and preserve the equality, as follows:
x^2-(3/2)x+(3/4)^2= 1/2 + (3/4)^2
Factor the left side to get:
[x-(3/4)]^2= 17/16
Take the square root of both sides to get:
x-(3/4) = (sqrt17)/4)or x-(3/4)= -(sqrt17)/4
x=[3+sqrt17]/4 or x=[3-sqrt17]/4
Cheers,
Stan H.
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