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
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
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
In general, if you have $\displaystyle x^2+bx=c\,,$ you should change b to $\displaystyle 2\left(\frac{b}{2}\right)$, and add $\displaystyle \left(\frac{b}{2}\right)^2$ to both sides of the equation. Then you have a perfect square on the left-hand side. (You have completed the square.)$\displaystyle x^2+2\left(\frac{b}{2}\right)x+\left(\frac{b}{2} \right)^2=\left(x+\frac{b}{2}\right)^2$
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)
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?
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!
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!!!
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 $\displaystyle x^2+4x+4 = (x+2)(x+2) = (x+2)^2$ and of course this is equal to 9.
Since you have $\displaystyle (x+2)^2 = 9$ you can take the square root: $\displaystyle x+2 = \pm 3$
Slightly different approach/explanation:
Literally, drag everything onto one side:
$\displaystyle x^2+4x-5=0$
Halve the x coefficient, and bracket it:
$\displaystyle (x+2)^2$
Now, if we expand this, we get $\displaystyle x^2+4x+4$ which is not what we had originally. So we need to subtract $\displaystyle 4$ and $\displaystyle 5$ to get what we had originally, and this gives:
$\displaystyle (x+2)^2-4-5=0$
Then, we can rewrite the "spare" terms on the right:
$\displaystyle (x+2)^2=9$
And then by taking the square root of both sides, we're there!