First on a fuction f(x)=2x^2-3 How do I find x such that f(x)=x How do I factorise 3x^2y-6xy+12xy^2 And solve the equation 1-3y=2(y-8)
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Originally Posted by t=0m First on a fuction f(x)=2x^2-3 How do I find x such that f(x)=x You are looking for a value of x such that $\displaystyle 2x^2 - 3 = x$ or $\displaystyle 2x^2 - x - 3 = 0$ See what you can come up with. (Hint: the quadratic factors.) -Dan
Originally Posted by t=0m How do I factorise 3x^2y-6xy+12xy^2 All three terms contain a common 3xy. So: $\displaystyle 3x^2y-6xy+12xy^2 = 3xy(x - 2 + 4y)$ is the best we can do. -Dan
Originally Posted by t=0m And solve the equation 1-3y=2(y-8) Come now. If you are supposed to be able to do the other two you should be able to do this one. I'll start you off: $\displaystyle 1-3y=2(y-8)$ $\displaystyle 1 - 3y = 2y - 16$ Can you take it from here? -Dan
The funtion is the top one and you need to make The number you put in (x) = x
Originally Posted by topsquark You are looking for a value of x such that $\displaystyle 2x^2 - 3 = x$ or $\displaystyle 2x^2 - x - 3 = 0$ See what you can come up with. (Hint: the quadratic factors.) -Dan Originally Posted by t=0m The funtion is the top one and you need to make The number you put in (x) = x Isn't that what I did? -Dan
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