MHB What are the values of x for f(x) = g(x)?

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To find the values of x for which f(x) = g(x), the equations f(x) = x^2 and g(x) = 10 - 3x are set equal to each other, resulting in the equation x^2 + 3x - 10 = 0. This quadratic equation can be solved using the quadratic formula or by inspection. The discussion emphasizes the importance of substituting the functions correctly to form a solvable equation. The participant expresses gratitude after understanding the solution process. The thread highlights a common challenge in solving equations by substitution and the collaborative nature of seeking help in math forums.
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Hello, mathhelpboards. I'm having trouble solving this problem on my review but can't figure out how to start it.

Find the values of x for which f(x) = g(x).

f(x) = x^2 , g(x) = 10 - 3x.

It seems like its easy but I can't really pick out what to put what where.
 
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Iwanttolearn said:
Hello, mathhelpboards. I'm having trouble solving this problem on my review but can't figure out how to start it.

Find the values of x for which f(x) = g(x).

f(x) = x^2 , g(x) = 10 - 3x.

It seems like its easy but I can't really pick out what to put what where.

Hey Iwanttolearn! Welcome to MHB! ;)

Let's see...
\[
f(x)=g(x) \Rightarrow x^2=10-3x \Rightarrow x^2+3x-10=0
\]
For which $x$ would that be the case? (Wondering)
 
I like Serena said:
Hey Iwanttolearn! Welcome to MHB! ;)

Let's see...
\[
f(x)=g(x) \Rightarrow x^2=10-3x \Rightarrow x^2+3x-10=0
\]
For which $x$ would that be the case? (Wondering)

So you would put the f(x) value into the g(x)? I think that looks right or either I'm completely wrong.
 
Iwanttolearn said:
So you would put the f(x) value into the g(x)? I think that looks right or either I'm completely wrong.

We're replacing the occurences of $f(x)$ and $g(x)$ by their respective definitions.
That gives us an equation we can solve with the quadratic formula, or just with inspection...
 
I like Serena said:
We're replacing the occurences of $f(x)$ and $g(x)$ by their respective definitions.
That gives us an equation we can solve with the quadratic formula, or just with inspection...

Oh wow! I get it now. Many thanks your way!(Handshake)(Happy)
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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