Hi guys basically I have got a bit of a problem with completing the square...
If I had to complete the square of x^{2}-14x+36=0
I would form the square of (x-7)^{2}
I would then expand the brackets to get x^{2}-14x+49
I would then combine this with the original equation to get x^{2}-14x+49+36=0
I then tidy it up to get x^{2}-14x+85=0
I then take 85 from both sides to leave me with x^{2}-14x=-85
My reaction to that would be to try and remove the ^{2} by square rooting each side but then I would be rooting a negative? Can someone give me some help by talking me through what to do as opposed to just showing me a page of maths? Sorry but I just don't learn particularly well that way.
Thanks in advance for anyone who helps me, it is appreciated.
Yes, that's correct.
This is where you went wrong. You cannot just add "49" to one side. You CAN, as Plato says, add 49 to both sides of the equation to getI would then combine this with the original equation to get x^{2}-14x+49+36=0
The other thing you can do is, noting that 36= 49- 13, write this as , then write it as .
It should be clear that this is impossible! If x^{2}- 14x+ 36= 0, then x^{2}- 14x+ 85 cannot also be 0- it is obviously 36 moreI then tidy it up to get x^{2}-14x+85=0
And now you have undone the whole point of "completing the square"! The left side, x^{2}- 14x is NOT a "perfect square" so you cannot take the square root.I then take 85 from both sides to leave me with x^{2}-14x=-85
It is, of course, possible that completing the square correctly will give a negative number. For example, if you have x^{2}- 2x+ 4= 0, you would "complete the square" by writing 4 as 1+ 3: x^{3}- 2x+ 1+ 3= (x-1)^{2}+ 3= 0 which is the same as (x- 1)^{2}= -3. Now you could either (depending on the "level" of the course) say "There is no (real number) value of x that satisfies the equation so the equation has no solution" or say that so that the two solutions are [tex]x= 1\pm i\sqrt{3}[tex].My reaction to that would be to try and remove the ^{2} by square rooting each side but then I would be rooting a negative? Can someone give me some help by talking me through what to do as opposed to just showing me a page of maths? Sorry but I just don't learn particularly well that way.
Thanks in advance for anyone who helps me, it is appreciated.
To avoid getting into issue of 85 etc we may proceed as under:
x^2 - 14 x + 36 = 0
step - 1 get rid of constant from LHS. In this case by adding ( -36) on both sides we get
x^2 - 14 x = - 36
Step 2. Now add square of half the coefficient of x on both sides. in this case the coefficient of x is -14, half is -7 square of ( -7) = 49.
Thus we add 49 on both the sides of the equation.
x^2 - 14 x + 49 = 49 - 36
( x-7)^2 = 13