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Quadratic Simultaneous Equations Help: A Level Maths

liamlarner2020

can someone check over my working to see where I have made a slipup the solution says the answer is y=-3/5 when x= 4/5

Quadratic Simultaneous Equations:
Solve:
2x+y=1
x²+y²=1

2x+y=1
y=1-2x let this be the new equation

now we have:
y=1-2x
x²+y²=1

substituting y into the second equation
x²+(1-2x)²=1
x²+1-2x-2x+4x²=1
x²+4x²-2x-2x+1=1
x²+4x²-4x+1=1
x²+4x²-4x=0
5x²-4x=0

in the form ax²+bx+c=0
a=5, b=-4,c=0

putting this in the quadratic formula gives:
x=4/5 or x=0

now substituting the x values into:
x²+y²=1 gives:
0²+y²=1
0+y²=1
y²=1-0
y²=1
√1=y
√1=1
so y=1 when x=0

now substituting x=4/5 gives:
(4/5)²+y²=1
(16/25)+y²=1
y²=1-16/25
y²=9/25
y=√9/25
y=3/5

the problem I face is I get +3/5 but the answers in the textbook say -3/5 when x = 4/5 where have I gone wrong.

Reply 1

username5714630

I'm not sure where you went wrong.
If you sub in x = 4/5 into 2x + y = 1 you will get a y value of -3/5.
x²+y²=1 is a circle so there will be two y values mapped to x at 4/5 where as 2x + y = 1 is a linear graph.

Reply 2

unplayable

You should sub in your x solutions into 2x+y=1 in order to get your y solutions. By subbing in the x solutions into the quadratic equation, you gain extra solutions (for example, y^2 = 1 implies y = 1 AND -1, but clearly the -1 solution does not satisfy 2x+y = 1).

Also you don't need to use the quadratic formula for 5x^2 - 4x = 0, you should notice that this factorises nicely into x(5x-4) = 0