Course:MATH110/Archive/2010-2011/003/Math Forum/Just-In-Time Chapter 3

Questions about section 3.2

Example 5

Hi, I'm a bit unclear about an intermediate step in one of the examples. The question is:

Solver for $y$ in $y^{2} + 2 y + 2 = 0$

The question is worked down to $y = \frac{- 2 \pm \sqrt{- 4}}{2}$

$y = \frac{- 2 \pm 2 \sqrt{- 1}}{2}$

Final Answer: $- 1 \pm \sqrt{- 1}$

Why does the question change (square root of -4) into $2 \sqrt{- 1}$ ? How do I make this change? It seems like a necessary step I just don't see how/why I'm doing this.

Thanks, Bernadette

Bernadette, was this a typo or an actual mistake: you wrote square root of 4 and not -4 as in the textbook. This might have caused some confusions. In any case, I edited your post and repalced 4 by -4. Cheers, DavidKohler

And since I'm here, I don't think it is so much of a necessary step. You can say there are no real solutions right away when you see that the number in the square root is negative. What is done in this example is an introduction to complex numbers.

Questions about section 3.1

If x is the length of the square then the area would be $x^{2}$ . But the length is increased by 6 so the length is now $(x + 6)$ . Therefore the area will be $(x + 6)^{2}$ . The question notes that the area of the square will increase by 156. So if the area to begin with was $x^{2}$ then the area will now be $x^{2} + 156$ . That means your equation is: $(x + 6)^{2} = x^{2} + 156$

Then you just solve for x:

$(x + 6)^{2} = x^{2} + 156$

$x^{2} + 12 x + 36 = x^{2} + 156$

$x^{2} + 12 x - x^{2} = 156 - 36$

$12 x = 120$

$x = 120 / 12$

$x = 10$

Then you can check: If $x = 10$ the original area was $1 0^{2} = 100$ . If the length is increased by 6 or $(10 + 6) = 16$ then the area would be $1 6^{2} = 256$ which is 156 more than 100.

Hope that helps. =) Victoria