SOLUTION: I find this quite difficult, though I am gaining some level of understanding. I am unsure how to solve for the following 2 quadratic equations Please help Thanks 1. x2 – 2x

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I find this quite difficult, though I am gaining some level of understanding. I am unsure how to solve for the following 2 quadratic equations Please help Thanks 1. x2 – 2x      Log On


   

Question 276593: I find this quite difficult, though I am gaining some level of understanding. I am unsure how to solve for the following 2 quadratic equations Please help
Thanks

1. x2 – 2x – 13 = 0
2. 2x2 – 3x – 5 =0
These are the steps I need to take, just unsure how
a. move the constant term to the right side of the equation
b. multiply each term in the equation by four times the cofficient 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
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
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
Found 2 solutions by richwmiller, solver91311:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-2x%2B-13+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-2%29%5E2-4%2A1%2A-13=56.

Discriminant d=56 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+56+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+56+%29%29%2F2%5C1+=+4.74165738677394
x%5B2%5D+=+%28-%28-2%29-sqrt%28+56+%29%29%2F2%5C1+=+-2.74165738677394

Quadratic expression 1x%5E2%2B-2x%2B-13 can be factored:
1x%5E2%2B-2x%2B-13+=+1%28x-4.74165738677394%29%2A%28x--2.74165738677394%29
Again, the answer is: 4.74165738677394, -2.74165738677394. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-13+%29

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-3x%2B-5+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-3%29%5E2-4%2A2%2A-5=49.

Discriminant d=49 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--3%2B-sqrt%28+49+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-3%29%2Bsqrt%28+49+%29%29%2F2%5C2+=+2.5
x%5B2%5D+=+%28-%28-3%29-sqrt%28+49+%29%29%2F2%5C2+=+-1

Quadratic expression 2x%5E2%2B-3x%2B-5 can be factored:
2x%5E2%2B-3x%2B-5+=+2%28x-2.5%29%2A%28x--1%29
Again, the answer is: 2.5, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-3%2Ax%2B-5+%29

also
(x+1)* (2x-5) = 0

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


What you have is (at least to me) a rather unorthodox method of a process called "completing the square." Despite the very unfamiliar process, I have proven to myself that it actually works in the general case.

Start with the general form of the quadratic equation:



Now, let's follow your steps:

"a. move the constant term to the right side of the equation"



"b. multiply each term in the equation by four times the cofficient of the 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"

First you need to factor the expression on the LHS of the equation. Fortunately, this is a perfect square expression:



(Verification of that last intermediate step is left as an exercise for the student)

Now we can take the square root of both sides, remembering to consider both the positive and negative roots -- such consideration typically indicated in the resulting RHS.



By considering both positive and negative roots at this point, you can do steps e and f simultaneously:



And we achieve a result that is identical to the quadratic formula which is the solution to the general quadratic.

Let's do one of your problems now that we are certain that the process will work for every problem:



a:

b:

c:

d:

d1:

e&f:

Then you can simplify a little by recognizing that 56 is 4 times 14, so:



Let's see if this answer makes any sense. 3 squared is 9 and 4 squared is 16, so 3.75 is a pretty good rough guess for square root 14. 1 plus 3.75 is 4.75 and 1 minus 3.75 is -2.75. So look where the graph crosses the -axis:



Pretty close, I'd say. Let me know if you still need help.

John