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 June 21st, 2017, 12:53 AM #1 Newbie   Joined: Jun 2017 From: Cape Town Posts: 6 Thanks: 0 Puzzled by functions Hi guys, I'm new on the block and stumbling my way along geometry. I've attempted them, but it just doesn't seem right. I'd appreciate some feedback: Question 1: Suppose f is defined as f(x) = x(x+1), solve the inequality f(x) > 6: $\displaystyle x(x+1) > 6$ $\displaystyle x^2 + x > 6$ $\displaystyle x^2 + x - 6 > 0$ $\displaystyle (x-2)(x+3) > 0$ I then end up with x <-3 or x > 2 as the solution. This seems correct but I'm not 100% sure. Question 2: Suppose h is defined as h(x) = 3sqrt(x+4) + x, solve the equation h(x) - 2x = 0 I try to substitute 3sqrt(x+4) + x into h(x) but end up with an equation I struggle to solve. $\displaystyle 3\sqrt{x+4} + x - 2x$ $\displaystyle (3\sqrt{x+4})^2 + x^2 - (2x)^2$ $\displaystyle 9(x+4) +x^2 - 4x^2$ Am I on the right track here? Or did I mess up the substitution? Appreciate any feedback and pointers! Last edited by skipjack; June 21st, 2017 at 02:21 AM.
 June 21st, 2017, 03:00 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,057 Thanks: 1618 These are algebra problems. You answered question (1) correctly. (2). Note that if $x + 4 \geqslant 0$, $\sqrt{x + 4}$ specifies the non-negative square root of $(x + 4)$. The equation $3\sqrt{x + 4} + x - 2x = 0$ implies $3\sqrt{x + 4} = x$ so any real solution must satisfy $x \geqslant 0$. Squaring both sides gives $9(x + 4) = x^2$, so $x^2 - 9x - 36 = 0$, i.e. $(x - 12)(x + 3) = 0$. Hence $x = 12$ (which satisfies the original equation).
 June 21st, 2017, 08:57 AM #3 Newbie   Joined: Jun 2017 From: Cape Town Posts: 6 Thanks: 0 Hi Skipjack, Thank you for moving my post to the correct section and also for your answer. Really appreciate it, I get it now.
June 21st, 2017, 10:44 AM   #4
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Quote:
 Originally Posted by PuzzleBot $\displaystyle 3\sqrt{x+4} + x - 2x$ $\displaystyle (3\sqrt{x+4})^2 + x^2 - (2x)^2$ $\displaystyle 9(x+4) +x^2 - 4x^2$
The first two of those are not equal. The square of an expression in the form of a + b + c is not a^2 + b^2 + c^2. It is a^2 + b^2 + c^2 + 2ab + 2ac + 2bc. Making a = 1, b = 2, and c = 3, makes (a + b + c)^2 = 36. 1^2 + 2^2 + 3^2 does not equal 36. 1^2 + 2^2 + 3^2 + 2(1*2) + 2(1*3) + 2 (2*3) = 36.

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