Are my solutions to these quadratic questions correct? (6 questions)

Question

1. The side length of a square is 4m. When each side is increased by the same amount, the area of the square doubles. By how much did the side length increase?

$4^2 = 16$ = original area

$(4+x)^2 = 16^2$

$(4+x) = \pm 16$

$x = 12, -20$ but $-20$ is not a possible answer so the answer is $12$

2. Three sides of a rectangle have a combined length of $100 m$, and the area of the rectangle is $1200m^2$. Find the dimensions of the region

$2x_{1}+x_{2}$ = 100

$x_{1}x_{2} = 1200$

$x_{1} = \frac{100 - x_{2}}{2}$

$\frac{100 - x_{2}}{2} x_{2} = 1200$

$100x_{2}-x_{2}^{2} = 2400$

$x = 40, x = 60$ and those are the dimensions of the region

3. A rectangular box has dimensions of $2p, p,$ and $p + 2$ and its surface area is $36m^2$. Find the areas of the individual faces.

Surface area is $2(lw+lh+wh) = 36$

let $l = 2p, w = p, h = p+2$

Then the surface area is $2p^2+2p^2+4p+p^2+2p = 18$

$5p^2+6p = 18$

Using completing the square:

$p^2+\frac{6}{5}p = 18$

$(p+\frac{3}{5})^{2} = \frac{99}{25}$

** I was told to round to two decimal places**

$p+\frac{3}{5} = \pm\frac{\sqrt{99}}{5} \approx \pm\frac{9.95}{5}$

$p = \frac{6.95}{5}, \frac{-12.95}{5}$ We reject the negative so $p = 1.39$

Then $lh = 9.42, lw = 3.86, wh = 4.71$

4. Consider two different boxes. The first has a height of $b + 5$ and a square base with side length $b$. The second box has a height of $b+2$ and a square base with side length $b + 1$. If the two boxes have the same volume, find the value of $b$.

Volume of a box is $lwh$ but in this case we have $s^2h$ So we calculate the two volumes and set them equal to each other.

Box 1:

$h = b+5, s = b \implies s^2 = b^2$

$V_{1} = b^3+5b^2$

Box 2:

$h = b+2, s = b+1 \implies s^2 = b^2+2b+1$

$V_{2} = (b+2)(b^2+2b+1) = b^3+2b^2+b+2b^2+4b+2$

Setting $V_{1} = V_{2}$:

$ b^3+5b^2 = b^3+2b^2+b+2b^2+4b+2$

$b^2-5b-2=0$

Using the quadratic formula:

$b = \frac{5\pm\sqrt{33}}{2} \approx \frac{5\pm 5.74}{2}$

$b = \frac{10.74}{2} = 5.37$ and $b = \frac{-0.74}{2}$ We reject the negative answer.

5. A box with a square base and no lid is to be made from a square piece of paper by cutting squares from the corners and folding up the sides. The cutout squares are 5 cm on a side. If the volume of the box is 100 cm3, find the dimensions of the original piece of paper.

area of the base is $s^2$, then volume of the box is $lwh$ so volume is $5x^2$ since $5$ is the height.

so $5x^2 = 100$

$x = \pm 2 \sqrt{5}$ but we reject the negative answer. So the dimensions are: $l = 2 \sqrt{5}, w = 2 \sqrt{5}, h = 5$

5.In the figure below, the two radii differ by 1, and the area of the grey circle is 4/5 the area of the white circle. Find the radius of the white circle.

Let $r$ be the radius for the white circle and $r-1$ for the smaller.

then the area of the grey circle is 4/5 the area of the white circle so

$\pi (r-1)^2 = \frac{4}{5} (\pi r^2)$

$\frac{1}{5} \pi r^2 +2\pi r +\pi = 0$

I multiplied everything by 5 and divided everything by $\pi$ and got:

$r^2+10r+5 = 0$

Then using completing the square:

$r^2+10r+25 = 20$

$(r+5)^2 = 20$

$r+5 = \sqrt{20}$

$r = -5 \pm \sqrt{2}$

Answer 1

Your solutions to problems 3, 4, and 5 are correct.

The side length of a square is $4~\text{m}$. When each side is increased by the same amount, the side length of the square doubles. By how much did the side length increase?

The original area was $A = (4~\text{m})^2 = 16~\text{m}^2$. If that area is doubled, then the new area is $2 \cdot 16~\text{m}^2 = 32~\text{m}^2$, so you should have $$(4 + x)^2 = 32$$

Three sides of a rectangle have a combined length of $100~\text{m}$, and the area of the rectangle is $1200~\text{m}^2$. Find the dimensions of the rectangle.

You lost a factor of $2$ when you substituted into the area equation.

Let the side lengths of the rectangle be $x$ and $y$. Then we must solve the system of equations \begin{align*} 2x + y & = 100~\text{m}\\ xy & = 1200~\text{m}^2 \end{align*} Solving the first equation for $y$ yields $y = 100~\text{m} - 2x$. Substituting $100~\text{m} - 2x$ for $y$ in the second equation yields \begin{align*} x(100~\text{m} - 2x) & = 1200~\text{m}^2\\ 100x~\text{m} - 2x^2 & = 1200~\text{m}^2\\ -2x^2 + 100x~\text{m} & = 1200~\text{m}^2\\ x^2 - 50x~\text{m} & = -600~\text{m}^2\\ x^2 - 50x~\text{m} + 600~\text{m}^2 & = 0\\ (x - 20~\text{m})(x - 30~\text{m}) & = 0 \end{align*} which gives $x = 20~\text{m}$ or $x = 30~\text{m}$.

If $x = 20~\text{m}$, then substitution into the equation $2x + y = 100~\text{m}$ gives $y = 60~\text{m}$, which is valid since $(20~\text{m})(60~\text{m}) = 1200~\text{m}^2$.

If $x = 30~\text{m}$, then substitution into the equation $2x + y = 100~\text{m}$ gives $y = 40~\text{m}$, which is valid since $(30~\text{m})(40~\text{m}) = 1200~\text{m}^2$.

The difference of the radii of the two concentric circles is $1$. The area of the grey circle is $4/5$ the area of the area of the white circle. Find the radius of the white circle.

Both your proposed answers are negative since you made a sign error. \begin{align*} \pi(r - 1)^2 = \frac{4}{5}\pi r^2\\ \pi(r^2 - 2r + 1) = \frac{4}{5}\pi r^2\\ r^2 - 2r + 1 & = \frac{4}{5}r^2\\ \frac{1}{5}r^2 - 2r + 1 & = 0\\ r^2 - 10r + 5 & = 0\\ r^2 - 10r & = -5\\ r^2 - 10r + 25 & = 20\\ (r - 5)^2 & = 20\\ r - 5 & = \pm 2\sqrt{5}\\ r & = 5 \pm 2\sqrt{5} \end{align*} Since $16 < 20 < 25$, $4 < 2\sqrt{5} < 5$. Hence, $5 - 2\sqrt{5} < 5 - 4 = 1$. Therefore, $(5 - 2\sqrt{5}) - 1 < 0$, so we must discard the root $5 - 2\sqrt{5}$ since the grey circle would have a negative radius. Therefore, the white circle has radius $r = 5 + 2\sqrt{5}$.