Choosing and rejecting inequality

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Homework Statement


y = |(x-2)(x-4)| and y = 6 -2x

find the exact values for which these two equation equal each other


Homework Equations





The Attempt at a Solution



Right I got it down to

2 \pm \sqrt{2} and 4 \pm \sqrt{2} and

and I've sketched the graph, however how do i work out which values to reject and which to use?

Thanks :)
 
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You should have two cases: one for 2 < x < 4 and the other for x < 2 or x > 4. These two cases correspond to the intervals where (x - 2)(x - 4) is negative and positive, respectively.

For the case 2 < x < 4, I solved the quadratic equation x^2 - 8x + 14 = 0. My solutions were 4 +/- sqrt(2). Since 4 + sqrt is larger than 4, it doesn't meet the restriction that 2 < x < 4, so I would discard it.

Is that enough help?
 
yes, thankyou I think I see what your saying

I will do some pratice questions

Thanks :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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