Simply solve: fg(x) = 3x^(2)-6x+17 Last question on me stoopid Maths paper. Can't really think of what to do. Silly me. I think I know how to do it now. Thanks.
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Originally Posted by StephenPoco Simply solve: fg(x) = 3x^(2)-6x+17 Last question on me stoopid Maths paper. Can't really think of what to do. Silly me. I think I know how to do it now. Thanks. Solve for what If $\displaystyle fg(x)=0$ use the quadratic formula
Originally Posted by StephenPoco Simply solve: fg(x) = 3x^(2)-6x+17 Last question on me stoopid Maths paper. Can't really think of what to do. Silly me. I think I know how to do it now. Thanks. Hi StephenPoco, I'm not sure what fg(x) is, but I would set the expression = 0 and use the quadratic formula to solve. $\displaystyle 3x^2-6x+17=0$ $\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
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