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I need to solve the system

$$6x+y=3\tag 1 $$

$$x^2+y^2=16\tag 2$$

So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.

I get $$x^2+(3-6x)^2 = 16$$

which is $$x^2 + 36x^2-36x+9-16=0$$

which is $$37x^2-36x-7=0$$.

This i need to solve by completing the square.

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    If completing the square confuses you, you could use quadratic formula if you are allowed. – 2017-01-12
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    Ive been told to do it by completing the square. thanks though :) – 2017-01-12

2 Answers 2

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To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get $$37^2x^2-36\times37x-7\times37=0\ .$$ To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives $$(37x-18)^2=7\times37+18^2\ .$$ I'm sure you can finish it from here.

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    Thank you. When i was taught this i was told you turn it into this format – 2017-01-13
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    Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please? – 2017-01-13
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    The formula you gave in your comment is missing a square but I assume that's just a typo. – 2017-01-13
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    Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end. – 2017-01-13
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Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.

There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are

Given $ax^2+bx+c=0$,

  1. Note the values of $4a=148$ and $b^2=1296$
  2. Subtract $c$ from both sides: $37x^2-36x=7$
  3. Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$
  4. Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$
  5. The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$

From there the solution $x=\dfrac{18\pm\sqrt{583}}{37}$ is straightforward.