two numbers that multiply to equal 10, add to equal 77/12

[Guest]

I need help sloving for x for this question:

. . .(x - (1/x))² - 77/12(x - (1/x)) + 10 = 0

Let (x - (1/x)) = y, so:

. . .y² - 77/12y + 10

. . .(y. . . ) (y. . . )

But what two numbers multiply to equal 10 and add to equal the fraction of 77/12?

Thank you.

 

[stapel]

I'm going to guess that you mean the following:

. . . . .\(\displaystyle \L \left(x\,-\,\frac{1}{x}\right)^2\, -\,\frac{77}{12}\left(x\,-\,\frac{1}{x}\right) \,+\,10\,=\,0\)

If so, then the substitution is fine. But instead of factoring, why not use the Quadratic Formula? If you have to factor, however, try multiplying through first, to get:

. . . . .\(\displaystyle \L 12y^2\,-\,77y\,+\,120\,=\,0\)

You should quickly be able to find factors of (12)(120) = 1440 that add up to -77.

Eliz.

 

[soroban]

Re: two numbers that multiply to equal 10, add to equal 77/1

Hello, bittersweet!

First of all, you set it up "the hard way".
Second, you can get rid of fractions in an equation.

I need help solving for x for this question:

. . .\(\displaystyle \left(x\,-\,\frac{1}{x}\right)^2\,-\,\frac{77}{12}\left(x\,-\,\frac{1}{x}\right)\,+\,10\:=\:0\)

We want two numbers with a product of \(\displaystyle 10\) and a sum of \(\displaystyle \frac{77}{12}\)

We have: \(\displaystyle \,\begin{Bmatrix}xy\:=\:10 \\ x\,+\,y\:=\:\frac{77}{12}\end{Bmatrix}\)

The second equation gives us: \(\displaystyle \,y\:=\:\frac{77}{12}\,-\,x\;\) [a]

Substitute into the first equation: \(\displaystyle \,x\left(\frac{77}{12}\,-\,x\right) \;= \;10\;\;\Rightarrow\;\;\frac{77}{12}x\,-\,x^2\;=\;10\)

Multiply by 12: \(\displaystyle \,77x\,-\,12x^2\;=\;120\) . . . (Try to remember that!)

We have a quadratic: \(\displaystyle \,12x^2\,-\,77x\,+\,120\;=\;0\)

\(\displaystyle \;\;\)which factors: \(\displaystyle \,(4x\,-\,15)(3x\,-\,8)\;=\;0\;\) **

\(\displaystyle \;\;\)and has roots: \(\displaystyle \,x\:=\:\frac{15}{4},\;\frac{8}{3}\)

Substitute these into [a] and we get: \(\displaystyle \,y\:=\:\frac{8}{3},\;\frac{15}{4}\)

Therefore, the two numbers are: \(\displaystyle \L\,\frac{15}{4}\) and \(\displaystyle \L\frac{8}{3}\)

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Check
Product: \(\displaystyle \L\,\frac{15}{4}\,\times\,\frac{8}{3} \;= \;\frac{\not{15}^5}{\not{4}}\,\times\,\frac{\not{8}^2}{\not{3}} \;= \; 10\) . . . check!

Sum: \(\displaystyle \L\,\frac{15}{4}\,+\,\frac{8}{3}\;=\;\frac{45}{12}\,+\,\frac{32}{12}\;=\;\frac{77}{12}\) . . . check!

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**

Unless you are experienced at Factoring,
\(\displaystyle \;\;\)the Quadratic Formula is easier.

We have: \(\displaystyle \L\,12x^2\,-\,77x\,+\,120\;=\;0\)

Then: \(\displaystyle \L\,x\;=\;\frac{-(-77)\,\pm\,\sqrt{(-77)^2\,-\,4(12)(120)}}{2(12} \;= \;\frac{77\,\pm\,\sqrt{169}}{24}\)

and we have: \(\displaystyle \L\,x\;=\;\begin{Bmatrix}\frac{77\,+\,13}{24}\:=\:\frac{90}{24}\:=\:\frac{15}{4} \\.\\ \frac{77\,-\,13}{24}\:=\:\frac{64}{24}\:=\:\frac{8}{3}\end{Bmatrix}\)