# 2012 STEP 2 Q7

I reached the stage where you have to prove the expression for $\lambda_1$

I realised the use of $x.x = 1$ and I came to the below expression

$\lambda^2 (3-2\gamma -2\beta + \alpha) + \lambda(2\alpha - \beta-\gamma) + \alpha + \beta + \gamma - 3 = 0$

From looking at answers on TSR, the above has $(\lambda + 1)$ as a factor - how would you spot this? - If you can spot this the result follows easily from there...
In addition when I used the quadratic formula for the expression, why did I have to use the negative sign instead of the positive (in the $\pm$ part of formula) to get the answer, although we know $\lambda$ is positive - how would you know which sign to use?

### Try sketching it

If you think about it, what you are doing is finding the intersection of a line with a circle. One point at which this happens is $Y_{1}$ but there is another obvious point where the line in question intersects the circle. Carry on with this logic and you should be able to get why $\lambda = -1$ is going to be a solution, and why we would use the other answer of the quadratic instead of $\lambda = -1$. Also, sketching it helps.

Alternatively, if you have gotten to $\lambda = -1$ some other way and are just wondering why we don't take it as our answer, just try plugging it back in and seeing what vector you get back out.

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