SOLUTION: I have been trying to figure out these equations all weekend and am having a problem trying to not only solve these problems, but also the correct way to do them. I would apprecia

Question 280963: I have been trying to figure out these equations all weekend and am having a problem trying to not only solve these problems, but also the correct way to do them. I would appreciate any and all help in order to complete this assignment for the university. It says there are 6 steps involved in the proper formatting of these questions involving moving the constant term to the right side of the equation, multiplying each term by 4 times the coefficient of the x squared term, squaring the coefficient of the original x term, taking the square root of both sides, and then setting the left side of the equation equal to the positive square root of the number on the right side and then setting the left side of the equation equal to the negative square root of the number on the right side. I am completely lost and am hoping that if I give you the four problems, that you may be able to show me how each one is completed. Thank you very much.
(1) xsquared-2x-13=0
(2) 4xsquared-4x+3=0
(3) xsquared+12x-64=0
(4) 2xsquared-3x-5=0
Found 3 solutions by stanbon, richwmiller, Theo:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
It says there are 6 steps involved in the proper formatting of these questions involving
a) moving the constant term to the right side of the equation,
b) multiplying each term by 4 times the coefficient of the x squared term,
c) squaring the coefficient of the original x term,
d) taking the square root of both sides,
e) and then setting the left side of the equation equal to the positive square root of the number on the right side and
f) then setting the left side of the equation equal to the negative square root of the number on the right side.
--------------------------
Let's try one:
(1) x^2-2x-13=0
a) x^2 - 2x = 13
b)4x^2 - 8x = 52
c)"square the coefficient of the original x term" gives you 4
d)sqrt(4x^2-8x) = sqrt(52)
e)sqrt(4x^2+8x) = sqrt(52)
f)sqrt(4x^2+8x) = -sqrt(52)
============================
As far as I can see those steps get you nowhere.
Step "c" is particularly puzzling because you don't do anything
with the squared coefficient.
Good luck with the rest.
Cheers,
Stan H.
============================

(2) 4xsquared-4x+3=0
(3) xsquared+12x-64=0
(4) 2xsquared-3x-5=0

Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
Stanborn was very generous in helping you.
Some website rules to remember
One problem per submission
NO similar problems
limit 4 submissions daily
squared is written ^2
cubed ^3
x^2-2x-13=0
There are three common ways to solve quadratic equations
1-factor
2-complete the square
3-quadratic formula
This one can't be factored
You can use complete the square.
SOME of your 6 steps look like it might be an attempt to complete the square.
x^2-2x-13=0
get the constant on the right
x^2-2x=13
the coeficient of x^2 must be one. It is.
take the coefficient of x not x^2
-2
take half
-2/2
square it
1
add it to both sides
x^2-2x+1=14
(x-1)^2=14
take square root
x-1=sqrt(14)
x-1=-sqrt(14)

see below for approximate values
Quadratic formula works on ALL quadratic equations.

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=56 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 4.74165738677394, -2.74165738677394. Here's your graph:

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
If you have trouble following the letters, then jump down to where I begin solving the problems to see how it works with numbers.

Hopefully after looking at the numbers, you will understand the letters, but if you just understand the numbers you should be ok.

The Completing the Squares Method works on the following principle.

Standard form of a quadratic equation is ax^2 + bx + c = 0

You divide both sides of this equation by a to get:

x^2 + (b/a)x + (c/a) = 0

You subtract (c/a) from both sides of the equation to get:

x^2 + (b/a)x = -(c/a)

You divide (b/a) by 2 to get (b/2a) and you square it to get (b/2a)^2.

You work on the principle that:

(x + (b/2a))^2 = x^2 + (b/a)x + (b/2a)^2 to convert your equation to:

(x + (b/2a))^2 - (b/2a)^2 = (-c/a)

You add (b/2a)^2 to both sides of the equation to get:

(x + (b/2a))^2 = (-c/a) + (b/2a)^2

Your equation of:

x^2 + (b/a)x = -(c/a) is translated to:

(x + (b/2a))^2 = (-c/a) + (b/2a)^2

You then take the square root of both sides of this equation to get:

x + (b/2a) = +/- sqrt ((-c/a) + (b/2a)^2)

You than subtract (b/2a) from both sides of this equation to get:

x = -(b/2a) +/- sqrt ((-c/a) + (b/2a)^2)

I'll do your problems next and hopefully you will be able to follow.

Problem Number 1:

xsquared-2x-13=0

Since xsquared is the same as x^2, this equation is the same as:

x^2 - 2x - 13 = 0

Your a term is 1.
Your b term is -2.
Your c term is -13.

You divide both sides of this equation by your a term to get:

x^2 - 2x - 13 = 0
Your b/a term is now -2.
Your c/a term is now -13.

since a was equal to 1, b/a term and c/a term are the same as you b and c terms.

You add your c/a term to both sides of your equation to get:

x^2 - 2x = 13

Your b/a term is equal to -2.

You take 1/2 of your b/a term to get (b/2a) = -1

You square it to get (b/2a)^2 = 1

You add the (b/2a)^2 term to the right side of the equation to get:

x^2 - 2x = 13 + 1 = 14.

You take the square root of the left side of your equation plus the (b/2a)^2 term to get the square root of x^2 - 2x + 1 and then square it to get:

(x-1)^2 = x^2 - 2x + 1

You replace x^2 - 2x with (x-1)^2 on the left side of your equation to get:

(x-1)^2 = 14

You take the square root of both sides of this equation to get:

x-1 = +/- sqrt(14)

You add 1 to both sides of this equation to get:

x = 1 +/- sqrt(14)

Those are your roots.

A graph of your equation looks like this:

1 + sqrt(14) = 4.741657387

1 - sqrt(14) = -2.741657387

You can see from the graph that y = 0 at approximately these values confirming that we calculated correctly.

We'll now move on to problem 2.

Problem 2 equation is:

4xsquared-4x+3=0

Since xsquared is the same as x^2, this equation becomes:

4x^2 - 4x + 3 = 0

Your a term is 4
Your b term is -4
Your c term is 3

You divide both sides of this equation by the a term to get:

x^2 - x + 3/4 = 0

This is equivalent to x^2 - x + .75 in decimal format.

Your b/a term is -1.
Your c/a term is .75

You subtract your c/a term from both sides of the equation to get:

x^2 - x = -.75

You take 1/2 of your b/a term to get (b/2a) = -.5
You square your b/2a term to get (b/2a)^2 = .25

You add your (b/2a)^2 term to the right side of your equation to get:

x^2 - x = -.75 + .25 which makes your equation equal to:

x^2 - x = -.5

You take the square root of the expression on the left side of your equation plus your (b/2a)^2 term and square it to get:

x^2 - x + .25 = (x-.5)^2

Note that your (b/2a) term of -.5 is used here.

You replace your equation of:

x^2 - x = -.5 with:

(x-.5)^2 = -.5

You take the square root of both sides of your equation to get:

x-.5 = +/- sqrt(-.5)

You add .5 to both sides of this equation to get:

x = .5 +/- sqrt(-.5)

These would be your roots but the result is not a real number so your equation has no roots because the requirement is that the roots have to be real.

A graph of your original equation is shown below:

You can see that the graph never crosses the x-axis which means that the roots are not real.

If you solve the equation using the imaginary roots, you will see that the equation is true, but the requirement is that the roots be real so you would say that this equation has no roots with the implication being that this equation has no real roots.

If there is a square root of a negative number in the roots, then they are not real.

Now to your problem 3.

Your equation is:

(3) xsquared+12x-64=0

Since xsquared is the same as x^2, your equation becomes:

x^2 + 12x - 64 = 0

Your a term is equal to 1.
Your b term is equal to 12.
Your c term is equal to -64.

You divide both sides of your equation by your a term to get:

x^2 + 12x - 64 = 0

Your b/a term is equal to 12.
Your c/a term is equal to -64.

b/a and c/a are the same as b and c because the a term was equal to 1 to start with.

You add your c/a term to both sides of the equation to get:

x^2 + 12x = 64

1/2 of your b term is equal to (b/2a) = 6.

your b/2a term squared is equal to (b/2a)^2 = 36

You add (b/2a)^2 to the right side of your equation to get:

x^2 + 12x = 64 + 36 = 100

You take the square root of x^2 + 12x + 36 to get (x+6) and you square it to get:

x^2 + 12x = 100 becomes:

(x+6)^2 = 100

You take the square root of both sides of this equation to get:

x+6 = +/- 10

You subtract 6 from both sides of this equation to get:

x = -6 +/- 10

This makes x = 4 and x = -16

A graph of your original equation looks like this:

You can see that when y = 0, x = -16 and x = 4 confirming the roots have been calculated correctly.

Your problem number 4 is shown below:

Your equation is:

2xsquared-3x-5=0 which is the same as:

2x^2 - 3x - 5 = 0

Your a term is equal to 2
Your b term is equal to -3
Your c term is equal to -5

You divide both sides of your equation by your a term of 2 to get:

x^2 - (3/2)x - (5/2) = 0

Your b/a term is equal to 3/2.
Your c/a term is equal to 5/2.

You add your c/a term to both sides of the equation to get:

x^2 - (3/2)x = (5/2)

You take 1/2 of your b/a term to get (b/2a) = -(3/4).

You square your b/2a term to get (b/2a)^2 = (9/16).

You add your (b/2a)^2 term to the right side of the equation to get:

x^2 - (3/2)x = (5/2) + (9/16).

You take the square root of x^2 - (3/2)x + your (b/2a)^2 term to get:

square root of x^2 - (3/2)x + (9/26) = (x - (3/4)).

You square (x-(3/4)) and replace x^2 - (3/2)x on the left side of your equation with (x-(3/4))^2 to get:

(x - (3/4))^2 = (5/2) + (9/16)

You take the square root of both sides of this equation to get:

x-(3/4) = +/- sqrt ((5/2) + (9/16))

You add (3/4) to both sides of this equation to get:

x = (3/4) +/- sqrt ((5/2) + (9/16))

In decimal form, this becomes:

x = .75 +/- sqrt (3.0625)

You solve for x to get x = 2.5 and x = -1

A graph of your original equation looks like this:

You can see that the roots are at x = -1 and x = 2.5 confirming that we solved this problem correctly.

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