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By eliminating y from the equations y = x - 4, 2x^2 - xy = 8, show that x^2 + 4x - 8 = 0 - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 1
Question 8
By eliminating y from the equations y = x - 4, 2x^2 - xy = 8, show that x^2 + 4x - 8 = 0. Hence, solve the simultaneous equations y = x - 4, 2x^2 - xy = 8, giving... show full transcript
Worked Solution & Example Answer:By eliminating y from the equations y = x - 4, 2x^2 - xy = 8, show that x^2 + 4x - 8 = 0 - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 1
Step 1
By eliminating y from the equations
Answer
To eliminate y from the equations, we start by substituting y from the first equation into the second one:
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Substitute:
egin{align*} y & = x - 4
ext{Substituting into } 2x^2 - xy = 8: \ 2x^2 - x(x - 4) & = 8 \ 2x^2 - x^2 + 4x & = 8 \ x^2 + 4x - 8 & = 0 \ \text{This confirms the required equation.} \ \text{So we have shown that } x^2 + 4x - 8 = 0.
\end{align*}
Step 2
Hence, solve the simultaneous equations
Answer
Now we will solve the simultaneous equations:
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From the first equation:
[ y = x - 4 ]
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From the second equation:
[ 2x^2 - xy = 8 ]
Substitute y:
[ 2x^2 - x(x - 4) = 8 ]
This leads to:
[ 2x^2 - x^2 + 4x - 8 = 0 ]
Which simplifies to:
[ x^2 + 4x - 8 = 0 ]
We can solve this quadratic equation using the quadratic formula:
[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
Here, a = 1, b = 4, c = -8:
[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-8)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 32}}{2} = \frac{-4 \pm \sqrt{48}}{2} = \frac{-4 \pm 4\sqrt{3}}{2} = -2 \pm 2\sqrt{3} ]
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Substituting back to find y:
[ y = x - 4 ]
Therefore:
[ y = (-2 \pm 2\sqrt{3}) - 4 = -6 \pm 2\sqrt{3} ]
Thus, the solutions are:
( x = -2 + 2\sqrt{3}, y = -6 + 2\sqrt{3} ) and ( x = -2 - 2\sqrt{3}, y = -6 - 2\sqrt{3} )