## The Astounding Power of Area

### 3.4 Level 5 Quadratics = All of Them!

[This material appears in Section 2 of the full course on Quadratics.]

Here we go! Level 5 quadratics.

EXAMPLE: Solve $$3x^{2}+5x+1=9$$.

Answer: This is the first example we’ve encountered with a first term more complicated than just $$x^{2}$$. We could divided throughout by $$3$$ and solve instead the equation $$x^{2} + \frac {1} {3} x + \frac {1} {3} = 3$$ and use the box method – and it will work (try it!) – but we will be thick in the midst of fractions.

In the previous level we would have multiplied through by $$4$$ to make the middle coefficient even. But in this problem we have leading term $$3x^{2}$$ which, when multiplied by four gives $$12x^{2}$$, not a natural perfect square.

Let’s try dealing with $$3x^{2}$$ first and make it into a perfect square by multiplying through by $$3$$.

$$9x^{2}+15x+3=27$$

The first term is $$3x \times 3x$$ but the middle term is $$15x$$, which has an odd coefficient. To avoid fraction, let’s also multiply through by $$4$$.

$$36x^{2}+60x+12=108$$

This has kept the first term a perfect square – we have $$36x^{2}=6x \times 6x$$ – and has made the second term even. It seems we are set to go!

The box shows we would like the number $$25$$ to appear. Let’s add $$13$$ to both sides.

$$36x^{2}+60x+25=121$$

$$(6x+5)^{2}=121$$

$$6x+5=11$$    or    $$6x+5=-11$$

$$6x=6$$    or    $$6x=-16$$

$$x=1$$    or    $$x=-\frac{8}{3}$$

Success!

EXAMPLE: Solve $$7x^{2}-x+1=9$$.

Answer: Let’s multiply through by $$7$$ to make the leading term a square

$$49x^{2}-7x+7=63$$

and through by $$4$$ to make the second term even (and to preserve the square)

$$196x^{2}-28x+28=252$$

Subtract $$27$$ from each side and we’re good to go!

$$196x^{2}-28x+1=225$$

$$(14x-1)^2=225$$

$$14x-1=15$$    or    $$14x-1=-15$$

$$14x=16$$    or    $$14x=-14$$

$$x=\frac{8}{7}$$    or    $$x=-1$$

EXAMPLE: Solve  $$2x^{2}+3x-3=5$$.

Answer: Let’s multiply through by $$2$$ to make the first term a perfect square.

$$4x^{2}+6x-6=10$$.

The middle term is even, so it seems we are good to go.

But we see we are dealing with fractions nonetheless. Let’s multiply through by $$4$$ in any case and we’ll see the fractions are cleared away.

$$16x^{2}+24x-24=40$$.

Let’s add $$33$$ to each side.

$$16x^{2}+24x+9=73$$

$$(4x+3)^{2}=73$$

$$4x+3=\sqrt{73}$$    or    $$4x+3=-\sqrt{73}$$

$$4x=-3+\sqrt{73}$$    or    $$4x=-3-\sqrt{73}$$

$$x=\frac {-3+\sqrt{73}} {4}$$    or    $$x=\frac {-3-\sqrt{73}} {4}$$

The numbers weren’t pretty, but the method is straightforward.

The previous example illustrates …

THE ULTIMATE BOX METHOD: To solve an equation of the form $$ax^{2}+bx+c=d$$.

i) Multiply through by $$a$$ (to make the first term a perfect square)
ii) Multiply through by$$4$$ (to avoid fractions)
iii) Draw the square box

and off you go!

THE BOX METHOD WILL NEVER LET YOU DOWN! The symmetry of a square is our friend.

EXAMPLE: Solve $$-2x^{2}+3x+7=1$$.

Answer: Let’s multiply through by $$-2$$ and then by $$4$$. That is, let’s multiply through by $$-8$$.

$$16x^{2}-24x-56=-8$$

$$16x^{2}-24x+9=-8+9+56$$

$$(4x-3)^{2}=57$$

$$4x-3=\pm \sqrt57$$

$$x=\frac {3 \pm \sqrt57} {4}$$

Done!

Question: Was it dangerous to use the $$\pm$$ symbol here?

EXAMPLE: Solve $$11x^{2}-x+5=0$$.

Answer: Let’s multiply through by $$11$$and by $$4$$, that is, through by $$44$$.

$$484x^{2}-44x+220=0$$.

Subtracting $$219$$ from both sides gives

$$484x^{2}-44x+1=-219$$

$$(22x-1)^2=-219$$

But there is no number whose square is negative! The box method is telling us there is no solution to this equation!

Comment: Every example here has been crafted to have a solution, but this need not always be the case. For example:

$$x^{2}=9$$ has exactly two solutions.

$$x^{2}=0$$ has exactly one solution.

$$x^{2}=-9$$ has no real solutions.

The box method turns every quadratic into an equation of the form:

$$(something)^{2}=A$$.

If $$A$$ is positive, there will be two solutions; if $$A$$ is zero, there will be one solution; and if $$A$$ is negative, there will be no solutions (at least in the number system we’re currently working in).

EXERCISE 1: Solve:

a) $$3x^{2}+7x+5=1$$

b) $$5x^{2}-x-18=0$$

c) $$3x^{2}+x-2=2$$

d) $$2x^{2}-3x=5$$

e) $$10x^{2}-10x=1$$

f) $$2x^{2}-3x+2=0$$

g) $$2x^{2}=9$$

h)  $$4-3x^{2}=2-x$$

EXERCISE 2OPTIONAL CHALLENGE

We solved $$3x^{2}+5x+1=9$$ as our first example in this lesson, but only after rejecting two initial approaches.

i) DEAL WITH FRACTIONS:

Divide through by $$3$$ and consider the equation $$x^{2} + \frac {5}{3} x+ \frac {1}{3} = 3$$. Draw the box for this equation, full of fractions, and show that even with the fractions the answers $$x=1$$ and $$x=-\frac{8}{3}$$ emerge.

ii) DEAL WITH SQUARE ROOTS:

Think of $$3x^{2}$$ as $$\sqrt3 x\times \sqrt3 x$$ and draw the box for $$3x^{2}+5x+1=9$$ with square roots appearing on its sides. Deal with them and follow the box method all the way through to show that the answers $$x=1$$ and $$x=-\frac{8}{3}$$ again emerge.

No matter what route you choose to take, the box method will not let you down!

EXERCISE 3: ANOTHER OPTIONAL CHALLENGE

Fractions and other types of numbers are unavoidable in solving quadratics with non-integer coefficients.

Solve

$$\sqrt3 x^{2}-\frac{1}{\sqrt2}x+\frac{1}{10}=0$$

by making the box method work! (Start by multiplying through by $$\sqrt3$$  perhaps?)

EXERCISE 4: TRY THIS ONE FOR SURE

a) A rectangle is twice as long as it is wide. Its area is $$30$$ square inches. What are the length and width of the rectangle?
b) A rectangle is four inches longer than it is wide. Its area is $$30$$ square inches. What are the length and width of the rectangle?
c) A rectangle is five inches longer than its width. Its area is $$40$$ square inches. What are the dimensions of the rectangle?

EXERCISE 5: Solve the following quadratic equations:

a) $$v^{2}-2v+3=27$$

b) $$z^{2} +4z=7$$

c) $$w^{2}-6w+5=0$$

d) $$\alpha^{2}-\alpha+1=\frac{7}{4}$$

Also note that $$x = (\sqrt x)^{2}$$. Solve the following disguised quadratics.

e) $$x-6\sqrt x+8=0$$

f) $$x-2\sqrt x=-1$$

g) $$x+2\sqrt x-5=10$$ WATCH OUT! Explain why only one answer is valid for this problem.

h) $$3\beta – 2\sqrt\beta=7$$

i) $$2u^{4}+8u^{2}-5=0$$

EXERCISE 6:

a) Show that $$2(x-4)^{2}+6$$ is quadratic.

b) Solve $$2(x-4)^{2}+6=10$$.

c) Consider $$y=2(x-4)^{2}+6$$. What $$x$$-value gives the smallest possible value for $$y$$?

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