Ok, so this question requires some quadratic re-arrangement. First, multiply out the brackets of y=x(x-3) to get y=x^2-3x. Then, as everything in the second equation is a "squared number", you can square root it to get x+y=3. Then bring the x to the other side to get y=3-x. Substitute that into the first equation, so that is makes : 3-x=x^2-3x. Bring over the 3-x to the other side to form the quadratic 'x^2-2x-3'. Then solve that to get (x-3)(x+1). This gives X solutions of +3 and -1. Substitute each value into one of the equations ( I suggest x+y=3). This gives two final solutions. When x=3, y=0. When x=-1, y=4.

Reply 7

uchihaitachi

OP

2

Original post by an_atheist

so we now have x^4-5x^2=9 yes? Make it all equal to zero, x^4-5x^2-9=0. If we take x^2=x, then it becomes a quadratic, solve and then sub in x=x^2.

omg thanks so much!!!! this was bugging me so much

Reply 8

an_atheist

15

Original post by uchihaitachi

omg thanks so much!!!! this was bugging me so much

I expanded wrong. Please see my revised post.

Reply 9

uchihaitachi

OP

2

Original post by ErwinJ

Ok, so this question requires some quadratic re-arrangement. First, multiply out the brackets of y=x(x-3) to get y=x^2-3x. Then, as everything in the second equation is a "squared number", you can square root it to get x+y=3. Then bring the x to the other side to get y=3-x. Substitute that into the first equation, so that is makes : 3-x=x^2-3x. Bring over the 3-x to the other side to form the quadratic 'x^2-2x-3'. Then solve that to get (x-3)(x+1). This gives X solutions of +3 and -1. Substitute each value into one of the equations ( I suggest x+y=3). This gives two final solutions. When x=3, y=0. When x=-1, y=4.

thank you!! :biggrin:

well jealous couldn't figure it out myself

Reply 10

an_atheist

15

Original post by ErwinJ

Ok, so this question requires some quadratic re-arrangement. First, multiply out the brackets of y=x(x-3) to get y=x^2-3x. Then, as everything in the second equation is a "squared number", you can square root it to get x+y=3. Then bring the x to the other side to get y=3-x. Substitute that into the first equation, so that is makes : 3-x=x^2-3x. Bring over the 3-x to the other side to form the quadratic 'x^2-2x-3'. Then solve that to get (x-3)(x+1). This gives X solutions of +3 and -1. Substitute each value into one of the equations ( I suggest x+y=3). This gives two final solutions. When x=3, y=0. When x=-1, y=4.

And the solutions given are incorrect. x=3,y=0 is correct, but x=1,y=4 is not, as it doesn't fit the second equation.

Reply 11

Cosmocos

2

LOL I'm giving this a try and just got x=0 and y=+or-3

Reply 12

TARS

3

Original post by uchihaitachi

thanks

i get that but you end up with x^4 and i cant work further than that

You can just factorise it, given that x² is a factor of all terms on the LHS:

x²+ [x(x-3)]² = 9
x²+ (x² - 3x)²= 9
x²+ (x⁴ - 6x³ + 9x²) = 9
x⁴ - 6x³ + 10x² = 9
x²(x² - 6x + 10) = 9
x²(x² - 6x + 10) = 9
so x²(x² - 6x + 10) - 9 = 0

(edited 8 years ago)

Reply 13

ErwinJ

10

Original post by uchihaitachi

thank you!! :biggrin:

well jealous couldn't figure it out myself

No problem! Good luck with maths! I am in Year 12 taking maths but, upon thought, I probably wouldn't have been able to do this question last year, but now it seems a doddle!

Reply 14

ErwinJ

10

Original post by an_atheist

And the solutions given are incorrect. x=3,y=0 is correct, but x=1,y=4 is not, as it doesn't fit the second equation.

Correct, it appears I over-looked that. However my methodology still works. There is no need for there to be x^4 popping up in anybodies working out of this question.

(edited 8 years ago)

Reply 15

Cosmocos

2

https://www.wolframalpha.com/input/?i=%28y%3Dx%28x-3%29%29+and+%28y%5E2+%2B+x%5E2%3D9%29

Wolfram Alpha gives really dodgy solutions...

Reply 16

the bear

20

Original post by ErwinJ

Ok, so this question requires some quadratic re-arrangement. First, multiply out the brackets of y=x(x-3) to get y=x^2-3x. Then, as everything in the second equation is a "squared number", you can square root it to get x+y=3. Then bring the x to the other side to get y=3-x. Substitute that into the first equation, so that is makes : 3-x=x^2-3x. Bring over the 3-x to the other side to form the quadratic 'x^2-2x-3'. Then solve that to get (x-3)(x+1). This gives X solutions of +3 and -1. Substitute each value into one of the equations ( I suggest x+y=3). This gives two final solutions. When x=3, y=0. When x=-1, y=4.

you seem to be saying that since x² + y² = 9 it follows that

x + y = 3

:confused:

Reply 17

an_atheist

15

Original post by ErwinJ

Look carefully. It is -1 not 1. It works.

-1^2=1, yes? 4^2=16. 1+16 doesnt equal 9, so you are wrong

Reply 18

TARS

3

Original post by Cosmocos

https://www.wolframalpha.com/input/?i=%28y%3Dx%28x-3%29%29+and+%28y%5E2+%2B+x%5E2%3D9%29

Wolfram Alpha gives really dodgy solutions...

How is x = 0 and x = 3 dodgy?

Reply 19

Rifleboy123

2

Original post by ErwinJ

Ok, so this question requires some quadratic re-arrangement. First, multiply out the brackets of y=x(x-3) to get y=x^2-3x. Then, as everything in the second equation is a "squared number", you can square root it to get x+y=3. Then bring the x to the other side to get y=3-x. Substitute that into the first equation, so that is makes : 3-x=x^2-3x. Bring over the 3-x to the other side to form the quadratic 'x^2-2x-3'. Then solve that to get (x-3)(x+1). This gives X solutions of +3 and -1. Substitute each value into one of the equations ( I suggest x+y=3). This gives two final solutions. When x=3, y=0. When x=-1, y=4.

I'm pretty sure this is wrong. sqrt(x^2 + y^2) is not equal to x + y hence x+y is not 3

Help me solve this tricky GCSE question!!!

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