quadratic equations

bumblebee123

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Jan 3, 2018
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I've been trying to figure this question out for a while, but I've never come across one like this before. Can anyone help to explain how I would answer this question?

Question: a) express x^2 - 4x + 1 in the form (x + a )^2 + b

b) hence solve the equation x^2 - 4x + 1 = 0 giving your answer in the form p +- sqrt q where p and q are integers

I managed to figure out part a) which was ( x - 2 )^2 - 3

any help with part b) would be really appreciated :)
 

Dr.Peterson

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Nov 12, 2017
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You need to solve the equation (x - 2)^2 - 3 = 0.

I would expect you to have been given examples like this before being asked the question; but if not, think about adding 3 to both sides, then taking the square root of each side. The nice thing about this form is that x appears in only one place, so you can isolate it more or less the same way you do in a linear equation.

Give it a try, and show us what you do.
 

bumblebee123

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Jan 3, 2018
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( x - 2 )^2 = 3
x - 2 = sqrt 3
x = 2 + sprt 3

p = 2, q = 3

this is the correct answer! thank you, Dr.Peterson
 

Harry_the_cat

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( x - 2 )^2 = 3
x - 2 = sqrt 3 …. this should be +/- sqrt 3
x = 2 + sprt 3 …. this should be 2 +/- sqrt 3

p = 2, q = 3

this is the correct answer! thank you, Dr.Peterson
Note the question said "giving your answer in the form p +- sqrt q where p and q are integers". See the +/-.

Note: When solving say x^2 = 16 the answer is x = 4 or -4, since 4^2=16 and also (-4)^2 =16, so a pos and neg solution must be included.
 

bumblebee123

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Jan 3, 2018
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196
Note the question said "giving your answer in the form p +- sqrt q where p and q are integers". See the +/-.

Note: When solving say x^2 = 16 the answer is x = 4 or -4, since 4^2=16 and also (-4)^2 =16, so a pos and neg solution must be included.
oops, forgot about that, thanks!
 

HallsofIvy

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Jan 27, 2012
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4,878
"Question: a) express x^2 - 4x + 1 in the form (x + a )^2 + b"

This is called "completing the square". You should have learned, before you could be expected to do a problem like this, that \(\displaystyle (x+ a)^2= x^2+ 2ax+ a^2\).
(You can get that by actually doing the multiplication: \(\displaystyle (x+ a)^2= (x+ a)(x+ a)= x(x+ a)+ a(x+ a)= x^2+ ax+ ax+ a^2= x^2+ 2ax+ a^2\).)

Compare the first part of that, "\(\displaystyle x^2+ 2ax\)", to your "\(\displaystyle x^2- 4x\)". They will be exactly the same if \(\displaystyle 2a= -4\) which means that \(\displaystyle a= -2\). Yes, \(\displaystyle (x- 2)^2= x^2- 4x+ 4\). In order to "complete the square", to make \(\displaystyle x^2-4x\) a complete square, you need to add 4. Of course you can't just "add 4", that would change the value. But you can do the old trick of "add" and "subtract" the same thing so that you change the "form" but not the value: \(\displaystyle x^2- 4x+ 1= x^2- 4x+ 4- 4+ 1= (x^2- 4x+ 4)+ (-4+ 1)= (x- 2)^2- 3\). That is "in the form (x+ a)^2+ b" with a= -2 and b= -3.

For (b), write \(\displaystyle x^2- 4x+ 1= (x- 2)^2- 3= 0\). Adding 3 to both sides, \(\displaystyle (x- 2)^2= 3\). Taking the square root of both sides, \(\displaystyle x- 2= \pm\sqrt{3}\) so, adding 2 to both sides, \(\displaystyle x= 2\pm\sqrt{3}\).
 
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