Question

Solve (x-3)^2=5

Answer 1

FIrst, expand the left hand side by multiplying \[(x-3)(x-3)\]

Answer 2

Okay, then do you FOIL it?

Answer 3

Sure, if that's how you prefer to remember it. Each piece of the first group gets multiplied by each piece of the second group. Then you combine like terms.

Answer 4

take a square root of both sides. (Another way to do it)

Answer 5

If I'm doing this right, when you combine like terms, you're left with X^2-9=5?

Answer 6

No

Answer 7

Mmm...no.

\[(x-3)(x-3) = x*x + x*(-3) + (-3)*x + (-3)(-3) = x^2 -3x -3x + 9\]\[=x^2-6x+9\]

So we have
\[x^2-6x+9 = 5\]

Answer 8

Want to take a crack at solving that?

Answer 9

We could solve this by completing the square, or using the quadratic formula. Can't do it by factoring this time.

Answer 10

@turtlepenguin do you know how to solve quadratics by either of those methods?

Answer 11

Well, I'll do it with the quadratic formula for you.

Quadratic formula says if we have an equation of the form \[ax^2+bx+c = 0, \,a\ne0\]The solutions can be found as\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

We need to rearrange our equation slightly to get it in that form:
\[x^2-6x+9=5\]Subtract 5 from both sides\[x^2-6x+9-5=5-5\]\[x^2-6x+4=0\]\[a=1,\,b=-6,\,c=4\]

Answer 12

Now we plug those values into the formula:
\[x = \frac{-(-6)\pm\sqrt{(-6)^2-4(1)(4)}}{2(1)}\]\[x=\frac{6\pm\sqrt{36-16}}2\]\[x=3\pm\sqrt{5}\]

Answer 13

Note that we could have gotten here more directly via @SolomonZelman's suggestion:

\[(x-3)^2 = 5\]Take the square root of both sides:\[(x-3) = \pm\sqrt{5}\]Solve for \(x\)
\[x-3=\sqrt{5}\]\[x=3+\sqrt{5}\]And the other solution
\[x-3=-\sqrt{5}\]\[x=3-\sqrt{5}\]

Or written more compactly, \(x = 3\pm\sqrt{5}\)

Answer 14

I went with the longer route, because the fact you were asking for help solving it suggested that you didn't know how to do either route, and needed the practice.

Answer 15

My teacher is not a very good teacher. I feel really dumb right now.

Answer 16

But I get it now, so thank you.

Answer 17

Well, you can't be expected to know it if you've never learned it! Would you like me to do it by completing the square, too?

Answer 18

Then you can compare the different approaches on the same problem.

Answer 19

Tell that to my chemistry teacher. I think I get completing the square a little, but if you wouldn't mind, yes please?

Answer 20

Okay, when we complete the square, we're trying to get back to the very form we had when we started this problem :-) But we'll pretend we were never there.

\[x^2-6x+9=5\]To complete the square, we want to find some value of \(a\) such that \[(x+a)^2 = x^2+2ax + a^2\]gives us the the first two terms, \(x^2-6x\)

Answer 21

So we take half of the value of the coefficient (number in front) of \(x\), square it, and add that to both sides of the equation. Because we are adding the same thing to both sides, we don't unbalance the equation — it's like hanging an equal weight on each end of the teeter-totter :-)

We have \[x^2-6x+9 = 5\]We take half of \(-6\), which is \(-3\), square it \((9)\), and add to both sides:
\[x^2 - 6x + 9 + 9 = 5 + 9\]
Now we can take the left 3 terms (in brackets)
\[[x^2-6x+9] + 9 = 5 + 9\]and replace it with \((x-3)^2\), where the \(-3\) is gotten from taking half of the coefficient of \(x\):
\[(x-3)^2 + 9 = 5 + 9\]Now we shuffle those numbers together on the right hand side:
\[(x-3)^2 + 9 - 9 = 5 + 9 - 9\]\[(x-3)^2 = 5\]
and then we would take the square root of both sides and solve for \(x\):
\[(x-3) = \pm\sqrt{5}\]\[x-3+3 = \pm\sqrt{5} +3\]\[x=3\pm\sqrt{5}\]

Answer 22

I think completing the square method is a lot easier.

Answer 23

Thanks!!

Answer 24

Completing the square gets messy when you have a number in front of the \(x^2\) term. At that point, I reach for the quadratic equation. Or if it's a real-life problem, and the coefficients aren't all nice round numbers...

Answer 25

Good to know both! And then there's also factoring, which we couldn't do here. I'll do a quick sample:

\[x^2-3x-10 = 0\]We need a pair of numbers that when multiplied give us -10, and when added give us -3. -5 and 2 is such a pair, so
\[x^2-3x-10=(x-5)(x+2) = 0\]
\[x-5=0\rightarrow x=5\]\[x+2=0\rightarrow x=-2\]

If we completed the square:
\[x^2-3x-10 =0\]\[x^2-3x-(\frac{3}2)^2 - 10 = (\frac{3}2)^2\]\[(x-\frac{3}2)^2-10= (\frac{3}2)^2\]\[(x-\frac{3}2)^2 = 10+\frac{9}4 = \frac{49}4\]Square root of both sides\[x-\frac{3}{2} = \pm\sqrt{\frac{49}4}\]\[x-\frac{3}{2} = \pm\frac{7}2\]\[x = \frac{3}2\pm\frac{7}2\]
\[x = 5,\,x=-2\]

or quadratic:
\[a=1, b = -3, c = -10\]
\[x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-10)}}{2(1)} = \frac{3\pm\sqrt{49}}{2} = \frac{3\pm7}{2}\]
Same answer.

Answer 26

Now you ought to be "cooking with gas" as my old man likes to say :-)

Answer 27

Thank you SO much, can I ask you another question, just to verify that I did this problem correctly?

Answer 28

sure, then I need to go...

Answer 29

Find the zeros of the function by factoring:

f(x)=x^2-3x+2
x^2-3x+2=0
(x-2)(x-1)
x=(2,1)

Answer 30

Well, easy way to check the zeros is to plug them into the original equation and make sure that you get 0 for the result:

x = 1:
\[(1)^2-3(1) + 2 = 1-3+2 = 3-3 = 0\checkmark\]
x=2:
\[(2)^2-3(2)+2 = 4-6+2 = 6-6 = 0\checkmark\]

Good job!

Answer 31

Thanks!

Answer 32

That should be "easy, as in easy to explain", not always "easy to do" if the polynomial is complicated and the zeros are awkward values :-)

Answer 33

Okay, now you can go, thanks for helping me!!!

Answer 34

As you can imagine, if one of the zeros turned out to be something like \(\dfrac{119}{377}\) that would not be a fun procedure in most cases!

Answer 35

In that case I just would call my cousin and ask her for help.

Answer 36

Or cry bc math is my enemy