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June 21st, 2017, 12:53 AM   #1
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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.
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June 21st, 2017, 03:00 AM   #2
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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).
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June 21st, 2017, 08:57 AM   #3
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Hi Skipjack,

Thank you for moving my post to the correct section and also for your answer. Really appreciate it, I get it now.
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June 21st, 2017, 10:44 AM   #4
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Quote:
Originally Posted by PuzzleBot View Post
$\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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