Part 1: Solve each of the quadratic equations below. Show your work. (3 points)

x2 – 36 = 0 and x2 = 8x – 12

Part 2: Describe what the solution(s) represent to the graph of each. (2 points)

Part 3: How are the graphs alike? How are they different? (2 points)

Question

Part 1: Solve each of the quadratic equations below. Show your work. (3 points)

x2 – 36 = 0 and x2 = 8x – 12

Part 2: Describe what the solution(s) represent to the graph of each. (2 points)

Part 3: How are the graphs alike? How are they different? (2 points)

jim766 · Answer

x^2 - 36 = 0    you can solve this one by getting the x^2 by itself
        +36    + 36

x^2 = 36              now take the sq rt of both sides

jim766 · Answer

what number times itself = 36

whpalmer4 · Answer

For $x^2=8x-12$ or $x^2-8x+12 = 0$ you can solve with the quadratic equation, by factoring, or by completing the square.  Factoring is probably the easiest.

anonymous · Answer

How do you describe the solution and how are the alike/different?

whpalmer4 · Answer

For the first one, be aware that there are two answers, not one, because there are two numbers which when multiplied by themselves give 36...

anonymous · Answer

6 is one of them

whpalmer4 · Answer

Yes.  And the other is very similar :-)

whpalmer4 · Answer

What do you get if you square a negative number?

anonymous · Answer

-6?

whpalmer4 · Answer

Right!  -6 is the other solution to $x^2=36$

anonymous · Answer

so how would I write that?

whpalmer4 · Answer

$x = \pm 6$ would be one way.  Or $x=6, x=-6$

anonymous · Answer

Math is not my subject I don't understand it and its killing me. I have to get my online math class finished by sunday

whpalmer4 · Answer

Okay, let's solve the other one, because I have to leave shortly.  Do you know how to do any of the 3 methods I mentioned?

anonymous · Answer

not really

anonymous · Answer

I still dont understand how to write them out as solved and part 2 or 3

whpalmer4 · Answer

Okay, let's do the quadratic, then, because it always works, even if it might be more work sometimes.

If you have a quadratic equation $ax^2+bx+c=0$, with $a\ne0$, then the solutions are given by the quadratic formula:$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$and it is just a matter of plugging in the numbers and doing the arithmetic.  For that first equation, which we've already solved, we had $x^2 - 36 = 0$ which we can rewrite as $1x^2 + 0x -36 = 0$ and so $a = 1, b = 0, c = -36$
Plugging into the formula, that gives us$$x = \frac{-(0)\pm\sqrt{(0)^2-4(1)(-36)}}{2(1)}$$and that simplifies to $$x=\frac{\pm\sqrt{144}}{2} = \frac{\pm12}{2} = \pm 6$$

whpalmer4 · Answer

For the second equation, we have to rearrange it a little bit to put it in that form:
$$x^2 = 8x – 12$$Subtract 8x from both sides$$x^2-8x=\cancel{8x}-\cancel{8x}-12$$Add 12 to both sides$$x^2-8x+12=0$$Now we see that $a = 1, b=-8, c = 12$ and I'll let you plug them into the formula.

anonymous · Answer

how are they alike /different?

whpalmer4 · Answer

What do the solutions represent?  Well, a picture may be worth a few words here:

This is the graph of $y = x^2-36$

anonymous · Answer

it represents 6, -6

whpalmer4 · Answer

You should make a graph of the other equation (\y = x^2-8x+12\) for comparison.

whpalmer4 · Answer

You're on the right track — the point where the curve crosses the x-axis are the solutions to the equation.

whpalmer4 · Answer

the points, in this case

anonymous · Answer

-33?

whpalmer4 · Answer

No, -33 isn't a solution to either of those equations.

whpalmer4 · Answer

How did you get that?

anonymous · Answer

on the picture at the bottom it curves at -33 on the y axis

whpalmer4 · Answer

x-axis, not y-axis, and those ticks are 2, not 1 (count the ticks between 0 and -10)

anonymous · Answer

okay so nevermind on that one.

whpalmer4 · Answer

You were asked to solve $x^2-36=0$.  What I graphed was $y = x^2-36$.  At the points where $y = 0$, we are at a value of $x$ such that $y = 0 = x^2-36$.  Therefore, those points are the solutions to $x^2-36=0$.  Does that make sense?

whpalmer4 · Answer

Why don't you try solving the other equation so I can check your work?  I have to leave in a few minutes.

anonymous · Answer

where do I use a b and c?

whpalmer4 · Answer

In the formula I gave you...

anonymous · Answer

For the second equation, we have to rearrange it a little bit to put it in that form:
x2=8x–12
Subtract 8x from both sides
x2−8x=8x−8x−12
Add 12 to both sides
x2−8x+12=0
Now we see that a=1,b=−8,c=12 and I'll let you plug them into the formula.

anonymous · Answer

this is the formula

anonymous · Answer

add 12 to to 12 and 0?

anonymous · Answer

so 12 and 24?

whpalmer4 · Answer

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

sorry, gotta go!