nosnip
  • nosnip
How to I factorise/solve this equation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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nosnip
  • nosnip
\[y ^{2} - x - 6x ^{2}\]
nosnip
  • nosnip
= 0
nosnip
  • nosnip
Do you know what type of equation it even is so I can google it?

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More answers

UnkleRhaukus
  • UnkleRhaukus
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
nosnip
  • nosnip
ah okay that's I wasn't sure if it was a quadratic :)
UnkleRhaukus
  • UnkleRhaukus
(i'm assuming there is no other information give about y)
nosnip
  • nosnip
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
nosnip
  • nosnip
\[y - 3x + 2 = 0\] \[y ^{2} - x - 6x ^{2} = 0\]
UnkleRhaukus
  • UnkleRhaukus
so you have two equations?
UnkleRhaukus
  • UnkleRhaukus
What do you get if you solve the first one \[y-3x+2=0\]for \(y\)?
nosnip
  • nosnip
y = 3x + 2
UnkleRhaukus
  • UnkleRhaukus
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 \(x\)
UnkleRhaukus
  • UnkleRhaukus
(after simplifying)
nosnip
  • nosnip
okay give me a second :)
nosnip
  • nosnip
Do you get the quadratic \[3x ^{2} + 11x + 4 = 0\]
UnkleRhaukus
  • UnkleRhaukus
almost, but the middle term isn't right
nosnip
  • nosnip
I'm not sure what else it could be?
UnkleRhaukus
  • UnkleRhaukus
What did you get for the expansion of \[(3x-2)^2=\]
nosnip
  • nosnip
\[9x ^{2} + 12x +4\]
UnkleRhaukus
  • UnkleRhaukus
\[(a-b)^2=a^2-2ab+b^2\]
nosnip
  • nosnip
I'm still not seeing where I went wrong sorry!
UnkleRhaukus
  • UnkleRhaukus
the +12 should be -12
nosnip
  • nosnip
ooh yes sorry you're right okay I can do it myself from there thankyou :)
nosnip
  • nosnip
I've got to go out but thatnlyou for your help I will give it a go later
UnkleRhaukus
  • UnkleRhaukus
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

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