Question

How to I factorise/solve this equation?

Answer 1

\[y ^{2} - x - 6x ^{2}\]

Answer 2

= 0

Answer 3

Do you know what type of equation it even is so I can google it?

Answer 4

you can rearrange it

\[y^2−x−6x^2 = 0\]
\[0 = 6x^2+x-y^2\]

so it looks like a quadratic,
where \(a= 6\), \(b=1\), \(c=-y^2\)

solve with the quadratic formula

Answer 5

ah okay that's I wasn't sure if it was a quadratic :)

Answer 6

(i'm assuming there is no other information give about y)

Answer 7

well it's part of a simultaneous equation? I'm confused because it's a non-calculator question and this seems difficult without a calculator

Answer 8

\[y - 3x + 2 = 0\]
\[y ^{2} - x - 6x ^{2} = 0\]

Answer 9

so you have two equations?

Answer 10

What do you get if you solve the first one
\[y-3x+2=0\]for \(y\)?

Answer 11

y = 3x + 2

Answer 12

substitute this into the second equation
\[y^2−x−6x^2=0\\
\downarrow\\
(3x + 2)^2-x-6x^2 =0\]

This is a quadratic equation in \(x\),

solve for both values of \(x\)

Answer 13

(after simplifying)

Answer 14

okay give me a second :)

Answer 15

Do you get the quadratic \[3x ^{2} + 11x + 4 = 0\]

Answer 16

almost, but the middle term isn't right

Answer 17

I'm not sure what else it could be?

Answer 18

What did you get for the expansion of
\[(3x-2)^2=\]

Answer 19

\[9x ^{2} + 12x +4\]

Answer 20

\[(a-b)^2=a^2-2ab+b^2\]

Answer 21

I'm still not seeing where I went wrong sorry!

Answer 22

the +12 should be -12

Answer 23

ooh yes sorry you're right okay I can do it myself from there thankyou :)

Answer 24

I've got to go out but thatnlyou for your help I will give it a go later

Answer 25

you should now use the quadratic formula
on the right (simplified) quadratic equation

to get two nice values of x,
then back substitute into equation 1, to get the corresponding y's