Square root functions, solving simulatenously: L(x) = J(root(K-x))

[CrypticSalter]

Hi everyone, I'm a first time poster and could really use some quick help!

I have 2 questions which I am struggling with and they are:
(a) If the square root function L(x) = J(root(K-x)) passes through (4,3) and has an x-int at (11.5,0), find J and K.
(b) Prove that L(x) = root((69-6x)/5).

For a, I assume I need to solve simultaneously so I did:
equation 1: 3 = J(root(K-4))
equation 2: 0 = J(root(K-11.5))
The values I get from my CAS calculator is that K = 11.5 and J = 1.1... decimal...

But in b, doesn't J have to equal 1 if there is no coefficient in front of the root? Could someone please help me?

Thanks.

 

[stapel]

(a) If the square root function L(x) = J(root(K-x)) passes through (4,3) and has an x-int at (11.5,0), find J and K.
(b) Prove that L(x) = root((69-6x)/5).

For part (a), I assume I need to solve simultaneously so I did:
equation 1: 3 = J(root(K-4))
equation 2: 0 = J(root(K-11.5))
The values I get from my CAS calculator is that K = 11.5 and J = 1.1... decimal...

I can't speak to your calculator, but you might want to try using algebra instead. For Equation 2 to be true, you must have:

. . . . .\(\displaystyle J\, =\, 0\, \mbox{ or }\, \sqrt{\strut K\, -\, 11.5\,}\, =\, 0\)

Since the former would make Equation 1 false, then it must be the latter. And the only way the square root can be zero is for the argument also to be zero. Since you are given that the x-intercept is exactly at x = 11.5, then K must also equal 11.5. Then Equation 1 gives:

. . . . .\(\displaystyle 3\, =\, J\, \sqrt{\strut 11.5\, -\, 4\,}\, =\, J\, \sqrt{\strut 7.5\,}\)

. . . . .\(\displaystyle \dfrac{3}{\sqrt{\strut 7.5\,}}\, =\, J\, =\, \dfrac{3}{\sqrt{\strut \dfrac{15}{2}\,}}\, =\, \sqrt{\, \dfrac{18}{15}\,}\, =\, \sqrt{\strut \dfrac{6}{5}\,}\, =\, \dfrac{\sqrt{\strut 30\,}}{5}\)

Unfortunately, 30 is not a perfect square, so you are left with a decimal approximation.

But in b, doesn't J have to equal 1 if there is no coefficient in front of the root?

Why? Since you have a value for J which involves a radical, and since the form they have provided involves a radical, where is the issue?

What did you get when you plugged the (algegraic) exact form (rather than the calculator's decimal approximation) into the formula for L(x)?

Please show all of your steps. Thank you!

 

[CrypticSalter]

Thanks

Thanks heaps for that! Your explanation was very clear. For some reason I was assuming that J had to equal 1 since the form was L(X) = J . root(....) but I had a feeling that I had to do something else.

Could you redirect me to a site that explains how to simplify surds? I unfortunately missed Year 10 advanced maths where they got into that stuff. E.G. where you simplify 3 / (root(15/2)) ----> (root(30))/5.

Again, thank you for your time!

 

[stapel]

Could you redirect me to a site that explains how to simplify surds?

There are loads of sites that discuss how to simplify square roots (and other radicals); try here to get started.