Applying quadratic equations.

Question

So I realise this is quite an easy question, but for some reason I can't see the solution. So the question followed on from a previous question where we used a quadratic equation to find the dimensions of a right angle triangle.

This question is: Find the dimensions of all rectangles in which the area equals the perimeter + $3.5$, and in which the longer sides are twice the length of the shorter sides.

So my solution is:

side = $x$
length = $2x$

so then: length x width = length + length + width + width + $3.5 $

$2x*x = 2x + 2x + x + x + 3.5 $

$2x^2 = 6x + 3.5 $

$-2x^2 + 6x + 3.5 = 0$

Then I would factorise this to find the positive values of $x$ that can be then used to determine the length of the rectangle. I am struggling to factorise while one of the values is $3.5$. Is there a way to more easily conceptualise how to factorise with non whole values? thanks.

Answer 1

"Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "

Yeah, it's called completing the square or the quadratic formula.

If you have $ax^2 + bx + c = 0$ and $a \ne 0$ then.....

$ax^2 + bx + c = 0$

$x^2 + \frac ba = -\frac ca$

$x^2 + 2\frac b{2a} = - \frac ca$

$x^2 + 2\frac b{2a} + (\frac b{2a})^2 = (\frac b{2a})^2 - \frac ca$

$(x + \frac b{2a})^2 + \frac {b^2}{4a^2} -\frac {4ac}{4a^2}=\frac {b^2 - 4ac}{4a^2}$

$x + \frac b{2a} = \pm \sqrt {\frac {b^2 - 4ac}{4a^2}} = \frac {\sqrt{b^2 - 4ac}}{2a}$

$x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$.

So in your case $x = \frac {-6 \pm \sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $

$= \frac {-6 \pm \sqrt{36 + 28}}{-4}$

$= \frac {6 \mp \sqrt{64}}{4} $

$= \frac {6 \mp 8}{4} $

$-\frac 12$ or $\frac 72$. As $x > 0$ we have $x = \frac 72$ so

the sides are $\frac 72$ and $7$ and area is $\frac 72*7 = \frac {49}2 = 24\frac 12$ and the perimeter is $\frac 72 + \frac 72 + 7 + 7 = 21$.

And .... indeed, $24\frac 12 = 21 + 3.5$.

Answer 2

Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as $$4x^2-12x-7=(2x-3)^2-9-7=0\iff (2x-3)^2=4^2$$ so the longer size is $2x=4+3=7\:$ and the shorter size is $\:x=7/2$.

Answer 3

$$2x^2 - 6x - 3.5 = 0$$

I wondered if the $ac$ method would still work. It sort of can.

$ac =2(-3.5) = -7$ and $b=-6$.

We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.

So we replace $bx=-6x$ with $-6x = -7x + 1x$.

\begin{align} 2x^2 - 6x - 3.5 &= 2x^2 -7x + 1x - 3.5 \\ &= 2x(x-3.5) + 1(x-3.5) \\ &= (2x+1)(x-3.5) \\ \hline x &\in \{-0.5, 3.5\} \end{align}