Question

ok

Answer 1

subtract 4 from both sides
then square both sides
then solve the resulting quadratic

Answer 2

not quite - remember you need to square both sides - you have taken the square root of the right-hand-side instead of squaring it

Answer 3

/almost/\[(x-4)^2\ne x^2+16\]

Answer 4

\[(x-4)^2=(x-4)(x-4)=x(x-4)-4(x-4)=...\]

Answer 5

? - just continue expanding what I wrote above to get the correct expansion for \((x-4)^2\)

Answer 6

/almost/ again :)

you need to pay close attention to the signs

Answer 7

no - you had this before and I said that was not correct. maybe 3rd time lucky? :)

Answer 8

perfect! so, now we go back to the original equation we were trying to solve:\[x+2=(x-4)^2=x^2-8x+16\]

Answer 9

you now need to rearrange this to get a standard quadratic equation in the form:\[ax^2+bx+c=0\]

Answer 10

no, we were left with:\[x+2=x^2-8x+16\]you need to rearrange this so that you get some quadratic in x on the left-hand-side of the equals sign and a zero the the right-hand-side

Answer 11

so start collecting like terms

Answer 12

first step might be to swap the two sides to get:\[x^2-8x+16=x+2\]

then subtract x from both sides
then subtract 2 from both sides

what do you end up with?

Answer 13

not quite - plz try again...

Answer 14

perfect!

Answer 15

so we now have:\[x^2-9x+14=0\]you should be able to factorise this

Answer 16

/almost/ yet again :D

the expression you gave will give:\[(x+2)(x+7)=x^2+9x+14\]

Answer 17

we need a "-9x"

Answer 18

perfect! so we now have:\[(x-7)(x-2)=0\]which means x equals two possible values - do you know what they are?

Answer 19

you need to use the "zero property" rule which states something like:


if \(a\times b=0\) then either 'a' or 'b' must be zero

Answer 20

in your case a=x-7
and b=x-2

Answer 21

so you have either:\[x-7=0\]or:\[x-2=0\]do you understand this?

Answer 22

correct! so try each value of x in the original expression, which was:\[\sqrt{x+2}+4=x\]

Answer 23

and which values give you a valid equation

Answer 24

perfect!

just out of interest: 2 could also be considered a solution as normally the square root of something can be a positive OR negative number

Answer 25

i.e. for x=2, \(\sqrt{x+2}=\sqrt{2+2}=\sqrt{4}=\pm2\)

Answer 26

but in this case I think they ONLY want you to consider the positive square root value

Answer 27

gr8! glad we got there in the end! :)

Answer 28

yw - and thank YOU as well for persevering my friend! :)

Answer 29

:)