# Correct my Pre-Calculus work please?

PPlease explain how to do these problems. I got my test back and I'm trying to see what I did wrong so I can do better on the next test.

• Given $g(x)=\sqrt{x+5}$ find $g^{-1}$

$x= \sqrt{y+5}$, I squared both sides $$x^2=y+5, \quad -x^2+y+5(x), \quad x^3+xy+5x$$ I am sure this is wrong but I'm stuck

• Find side $A$ in a triangle with only the sides given: $6$, $15$, $17$. Also find the area of the triangle What I did: $$\cos A = (b^2 + c^2 - a^2)/2bc = (6^2 + 17^2 -15^2)/(2\cdot6\cdot 17) = (36+225-289)/180 = -0.155$$ so $A= \cos^{-1}(-0.155)$.

I am not sure how to find the area of this triangle.

• simplify $$\frac{\sqrt[3]{x^2y}^6}{x^{-3}y^5}$$ I squared the top by 6 and then tried to put like terms together. I ended up with: $$\frac{x^5 y^3}{x^{-3} y^5} = \frac{x^2 y^3 }{ y^5}$$

• Solve $x^2+12x=-27$

I did the diamond method and got $9x$ and $3x$ but couldn't factor it. I have alot of trouble with factoring. Like in the last problem $5x^5-125xy^4$ I didn't even know where to start. Please help me. Thank you

• latexify your text. i can't understand anything. – Mojojojo Oct 15 '13 at 10:42

1) $g(x) = \sqrt{x + 5} \Leftrightarrow g(x)^2 = x+5 \Leftrightarrow g(x)^2 - 5 = x$ for $x \geq -5$. So the inverse function is defined by $g^{-1}(x) = x^2 - 5$, on the domain $[0,\infty)$.

When finding an inverse to a function $g(x) =$ some expression in $x$, you should 'free' the $x$ on the right hand side and express it in terms of $g(x)$. I'm not sure what you did after squaring both sides, but it's not making much sense.

2) I don't know what you mean by "find side A". For the area, use Heron's formula. The area is equal to $\sqrt{S(S-a)(S-b)(S-c)}$ where $S = \frac{a+b+c}{2}$ and $a,b,c$ are the lengths of the three sides. Plugging in numbers gives $S = 19$ and area = $\sqrt{19 \cdot 13 \cdot 4 \cdot 2} = \sqrt{1976} = 2 \sqrt{494}$.

3) The numerator is equal to $(\sqrt[3]{x^2 y})^6 = ((\sqrt[3]{x^2 y})^3)^2 = (x^2 y)^2 = x^4 y^2$. The whole expression then simplifies to $\frac{x^4 y^2}{x^{-3}y^5} = x^7 y^{-3}$.

I'm not sure what you mean by 'squaring the top by 6', but what you did does not work. Also watch out at the last step: $x^2 \neq \frac{x^5}{x^{-3}} = x^{5-(-3)} = x^8$.

4) $x^2 + 12x = -27 \Leftrightarrow x^2 + 12x + 27 = 0 \Leftrightarrow (x+9)(x+3) = 0 \Leftrightarrow x = -9$ or $x = -3$.

When you factor a term of the form $x^2 + bx + c$, you want to find $p,q$ such that $p+q = b$ and $pq = c$. Then you can write $x^2 + bx + c = (x+p)(x+q)$. Work out the right hand side to convince yourself of this.

An alternative solution (completing the square): $x^2 + 12x = -27 \Leftrightarrow x^2 + 12x + 36 = -27 + 36 \Leftrightarrow (x+6)^2 = 9 \Leftrightarrow x+6 = \pm \sqrt{9}$
$\Leftrightarrow x+6 = \pm 3 \Leftrightarrow x = \pm 3 - 6 \Leftrightarrow x = -3$ or $x = -9$.