How to solve simultaneous equations with a quadratic
Example.
" Solve the simultaneous equations
y + 4x + 1 = 0
y^2 + 5x^2 + 2x = 0 "
We can rearrange the first equation to something that we can substitute into the second equation, for example
y = -4x - 1
We then substitute this into the second equation
(-4x - 1)^2 + 5x^2 + 2x = 0
16x^2 + 8x + 1 + 5x^2 + 2x = 0
21x^2 + 10x + 1 = 0
(7x + 1)(3x + 1) = 0
Gives us the solutions
x = -1/7 and x = -1/3
We then put these values into our first equation (y = -4x - 1) to give
y = -3/7 and y = 1/3
TW
Related Maths A Level answers
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If the quadratic equation kx^2+kx+1=0 has no real roots, what values of k are possible?
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Consider the function f (x) = (2/3) x^3 + bx^2 + 2x + 3, where b is some undetermined coefficient: (a) find f'(x) and f''(x) and (b) if you know that f(x) has a stationary point at x = 2, use this information to find b.