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Answers please

help.PNGi need help with these. answers please
Reply 1
Original post by ngunn
help.PNGi need help with these. answers please


Which methods have you learnt for solving quadratics? Have you done any working so far?

EDIT: Oops I just realised it says by bompleting the square!
(edited 10 years ago)
Reply 2
Original post by alow
Which methods have you learnt for solving quadratics? Have you done any working so far?

EDIT: Oops I just realised it says by bompleting the square!


i dont understand any of it
Reply 3
Original post by ngunn
i dont understand any of it


For the first one start by finding an expression of the form (x+a)2+c\left( x+a \right)^{2}+c where a and c are constants and is equal to x2+18x10x^{2}+18x-10

So, (x+a)2+c=x2+18x10\left( x+a \right)^{2}+c=x^{2}+18x-10

HINT: Find a by halving the coefficient of the xx term in the origional quadratic (which in the first question is 18).
Reply 4
are ok thanks
Original post by ngunn
i dont understand any of it


Don't you think you should speak to your teacher then?
Reply 6
Original post by alow
For the first one start by finding an expression of the form (x+a)2+c\left( x+a \right)^{2}+c where a and c are constants and is equal to x2+18x10x^{2}+18x-10

So, (x+a)2+c=x2+18x10\left( x+a \right)^{2}+c=x^{2}+18x-10

HINT: Find a by halving the coefficient of the xx term in the origional quadratic (which in the first question is 18).


can you give me the answer to the first one so i can see how it worked please
Original post by ngunn
can you give me the answer to the first one so i can see how it worked please

(x+a)2+c=x2+2ax18x+a2+c10=x2+18x10(x+a)^2+c=x^2+ \underbrace{2ax}_{18x} + \underbrace{a^2+c}_{-10}=x^{2}+18x-10. Hope that helped...
(edited 10 years ago)
Reply 8
Original post by ngunn
can you give me the answer to the first one so i can see how it worked please

I'll give you the start:

(x+a)2+c=x2+18x10a=9[br][br](x+9)2+c=x2+18x10[br][br]x2+19x+81+c=x2+18x10[br][br]\left( x+a \right)^{2}+c=x^{2}+18x-10 \Rightarrow a=9[br][br]\Rightarrow \left( x+9 \right)^{2}+c=x^{2}+18x-10[br][br]\Rightarrow x^{2}+19x+81+c=x^{2}+18x-10[br][br]

Can you see how to continue this and find the values of x? If not, as Mr M said, it's probably better to go to your teacher.
Reply 9
Original post by keromedic
(x+a)2+c=x2+2ax18x+a2+c10=x2+18x10(x+a)^2+c=x^2+ \underbrace{2ax}_{18x} + \underbrace{a^2+c}_{-10}=x^{2}+18x-10. Hope that helped...


how do i get the two possible values for x from that
Original post by ngunn
how do i get the two possible values for x from that

I don't know what you're asking, sorry.
Original post by ngunn
can you give me the answer to the first one so i can see how it worked please


That looks like it's MyMaths. Have you tried doing the lesson on there?
Original post by ngunn
how do i get the two possible values for x from that


Because if you initially brought all the constants over two one side, then you know that (x+a)2+c=0(x+a)^2+c=0 and you can solve this (remembering the +- when square rooting) to find the two possible values of x.
(edited 10 years ago)

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