Solving equations by completing the square.

**carrieh** · November 9th 2011, 09:46 AM

I am attempting to help my son with his homework and am completely stuck.. need to know how to solve the following:

x2+4x=5 and show workings can anyone please help and explain to me in very simple terms?

Thanks
Carrie

**e^(i*pi)** · November 9th 2011, 09:49 AM

Originally Posted by carrieh

I am attempting to help my son with his homework and am completely stuck.. need to know how to solve the following:

x2+4x=5 and show workings can anyone please help and explain to me in very simple terms?

Thanks
Carrie

What have you tried? Add 4 to both sides to get a perfect square on the left (and right for that matter)

$x^2 +4x + 4 = 5+4 \Longleftrightarrow (x+2)^2 = 9$

**Amer** · November 9th 2011, 09:56 AM

to complete the square you add half of x coefficient squared
for example

x^2 + 12 x - 13 , (12/2)^2 = 36
x^2 + 12 x + 36 - 36 -13
(x+6)^2 - 49

another one

x^2 + 5 x + 4
(5/2)^2 = 6.25

x^2 + 5x + 6.25 - 6.25 + 4
(x+ 2.5)62 - 2.25

**SammyS** · November 9th 2011, 09:58 AM

Originally Posted by carrieh

I am attempting to help my son with his homework and am completely stuck.. need to know how to solve the following:

x2+4x=5 and show workings can anyone please help and explain to me in very simple terms?

Thanks
Carrie

In general, if you have $x^2+bx=c\,,$ you should change b to $2\left(\frac{b}{2}\right)$ , and add $\left(\frac{b}{2}\right)^2$ to both sides of the equation. Then you have a perfect square on the left-hand side. (You have completed the square.)

$x^2+2\left(\frac{b}{2}\right)x+\left(\frac{b}{2} \right)^2=\left(x+\frac{b}{2}\right)^2$

**carrieh** · November 9th 2011, 10:01 AM

my equation is x^2+4x=5

I need to solve the equation by completing the square. Using your example i am stuck already, sorry.. did say simple terms!!! Is half of my coefficent squared - 0.5^2? which would be 0.25? or is it the coefficient of x not x^2? (baby steps i know! sorry)

**e^(i*pi)** · November 9th 2011, 10:03 AM

Originally Posted by carrieh

my equation is x^2+4x=5

I need to solve the equation by completing the square. Using your example i am stuck already, sorry.. did say simple terms!!! Is half of my coefficent squared - 0.5^2? which would be 0.25? or is it the coefficient of x not x^2? (baby steps i know! sorry)

The coefficient of x is the number next to x. In your question the coefficient of x is 4

**carrieh** · November 9th 2011, 10:14 AM

so in x^2+4x=5 My coefficient is (4/2)^2 =4

then don't understand where the 4 fits into my equation would it be x^2 + 4x + 4 -4 =5? Your equation does not show an = does this matter and does it affect the outcome?

**e^(i*pi)** · November 9th 2011, 10:17 AM

Originally Posted by carrieh

so in x^2+4x=5 My coefficient is (4/2)^2 =4

then don't understand where the 4 fits into my equation would it be x^2 + 4x + 4 -4 =5? Your equation does not show an = does this matter and does it affect the outcome?

Yes although you can move that -4 over to the other side and change it's sign

You should get $x^2+4x+4 = 9$ and since you have $x^2 + 2\left(\dfrac{4x}{2}\right) + 2^2$ you can factor it as per post 4 to get $(x+2)^2 = 9$

**carrieh** · November 9th 2011, 10:30 AM

ok so now i have

x^2 +4x +4 =9 (and i understand how we got there!)

the way my sons maths book shows how to work it out is this:

x^2 - 6x + 2 = 0
add -2 to each side
x^2 - 6x = -2
add 9 to each side
x^2 - 6x + 9 = -2+9 (I understand up until this part)
(x-3)^2 = 7 (I do not know how to get it to this? sorry!

**carrieh** · November 9th 2011, 10:35 AM

Have to run to take son canoeing, will be back about 9pm to check on answers and more than likely with more questions! Thanks so much for your help so far, i'm not the easiest to teach!!!

**e^(i*pi)** · November 9th 2011, 10:46 AM

Originally Posted by carrieh

ok so now i have

x^2 +4x +4 =9 (and i understand how we got there!)

the way my sons maths book shows how to work it out is this:

x^2 - 6x + 2 = 0
add -2 to each side
x^2 - 6x = -2
add 9 to each side
x^2 - 6x + 9 = -2+9 (I understand up until this part)
(x-3)^2 = 7 (I do not know how to get it to this? sorry!

They factorised the left hand side (does your son know how to factor a quadratic normally?) which was a perfect square. You can check by expanding (x-3)^2 using FOIL.

Note that your equation is also in the form of a perfect square but, if you prefer, which two numbers sum to 4 and multiply to 4? The answer is 2 and 2 so you have $x^2+4x+4 = (x+2)(x+2) = (x+2)^2$ and of course this is equal to 9.

Since you have $(x+2)^2 = 9$ you can take the square root: $x+2 = \pm 3$

**Quacky** · November 9th 2011, 11:01 AM

Slightly different approach/explanation:

Literally, drag everything onto one side:

$x^2+4x-5=0$

Halve the x coefficient, and bracket it:

$(x+2)^2$

Now, if we expand this, we get $x^2+4x+4$ which is not what we had originally. So we need to subtract $4$ and $5$ to get what we had originally, and this gives:

$(x+2)^2-4-5=0$

Then, we can rewrite the "spare" terms on the right:

$(x+2)^2=9$

And then by taking the square root of both sides, we're there!

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