### 2.3 Level Two, Three and Four Quadratics

Lesson materials located below the video overview.

Recall that our goal now is to recognize complicated expressions as simple level 1 quadratics in disguise.

EXAMPLE 26: Solve $$x^{2}+4x+4=25$$.

Answer: Let’s draw the square for $$x^{2}+4x+4$$.

There is an $$x^{2}$$  piece that must come from $$x \times x$$.

Because we want the figure to be a perfectly symmetrical square (squares are good for level 1) the “$$4x$$ ” must come from two symmetrical pieces: $$2x$$ and $$2x$$.

This means we must have the numbers $$2$$ and $$2$$ on the sides of the square, and this is consistent with the final portion being $$4$$.

Thus we see that $$x^{2}+4x+4$$ is really $$(x+2)^{2}$$ in disguise, and we need to solve:

$$(x+2)^{2} = 25$$.

This is easy:

$$x+2=5$$    or    $$x+2=-5$$

$$x=3$$    or    $$x=-7$$.  □

EXAMPLE 27: Solve $$x^{2}-10x+25=169$$.

Answer: Let’s draw the square for $$x^{2}-10x+25$$.

There is an $$x^{2}$$ piece that must come from $$x \times x$$ . And we need two (symmetrical) pieces that add to $$=10x$$.

This means the need the numbers $$-5$$ and $$-5$$, which is consistent with the final portion of the square being $$25$$.

$$x^{2}-10x+25=169$$

$$(x-5)^{2}=169$$

$$x-5=13$$    or    $$x-5=-13$$

$$x=18$$   or   $$x=-8$$   □

EXAMPLE 28: Solve $$x^{2}-20x+100=7$$.

$$x^{2}-20x+100=7$$

$$(x-10)^{2}=7$$

$$x-10=\sqrt 7$$    or   $$x-10=-\sqrt 7$$

$$x=\sqrt 7 + 10$$    or   $$x= -\sqrt 7 +10$$  □

EXAMPLE 29: Solve $$x^{2} +2 \sqrt 5 x + 5 = 4$$.

Answer: Don’t be perturbed by the numbers. Just follow things through as before.

Do you see we have this square?

$$x^{2} +2 \sqrt 5 x + 5 = 4$$

$$(x+\sqrt5)^{2} = 4$$

$$x+\sqrt5 = 2$$    or    $$x+\sqrt5 = -2$$

$$x=2-\sqrt5$$    or    $$x=-2-\sqrt5$$  □

Recall that all solutions to the practice problems appear in the COMPANION GUIDE to this QUADRATICS course.

PRACTICE 30: Solve:

a) $$x^{2}+40x+400=100$$

b) $$p^{2}-6p+9=9$$

c) $$x^{2}-4x+4=1$$

d) $$3x^{2}-18x+27=12$$

e) $$x^{2}-2\sqrt2 x+2=19$$

PRACTICE 31: Solve $$x^{2}+2Bx+B^{2}=D^{2}$$  in terms of $$B$$ and $$D$$.

EXAMPLE 32: Solve $$x^{2}+8x+15=80$$.

Answer: Let’s apply the technique of level 2 and draw the box.

The $$x^{2}$$ piece must come from $$x \times x$$. And because we want the symmetry of a square, $$8x$$ must come for two pieces $$4x$$each:

This means we must have the numbers $$4$$ and $$4$$ on the sides of the square, giving us $$16$$ for the remaining piece of the squares, WHICH IS INCONSISTENT WITH THE NUMBER $$15$$ in the problem.

It seems the box method is not of help. We have two options:

1. Give up and cry.
2. Be an adult and make it work!

Let’s follow option 2 making use of a good piece of general advice:

If there is something in life you don’t like and are not happy with, change it!

In this problem we would like the number $$15$$ in $$x^{2}+8x+15$$ to actually be a $$16$$. So let’s add one and make it $$16$$!

Of course, to keep the equation balanced, if we add 1 to one side of an equation we need to add 1 to the other side as well:

$$x^{2}+8x+15+1=80+1$$.

That is, we have:

$$x^{2}+8x+16=81$$.

And the box tells us that this is:

$$(x+4)^{2}=81$$

and all now falls into place:

$$x+4=9$$    or    $$x+4=-9$$

$$x=5$$     or    $$x=-13$$  □

EXAMPLE 33: Solve $$x^{2} -6x+11=27$$.

Answer: The box tells us that the $$x^{2}$$ and the $$-6x$$ pieces want the number $$9$$ to accompanying them, not $$11$$:

Let’s make that happen. Subtract $$2$$ from both sides:

$$x^{2}-6x+9=25$$

$$(x-3)^{2}=25$$

$$x-3=5$$    or    $$x-3=-5$$

$$x=8$$    or    $$x=-2$$  □

EXAMPLE 34: Solve $$x^{2}-10x+3=0$$.

Answer: The box tells us that the $$x^{2}$$ and the $$-10x$$ pieces want the number $$25$$ to accompanying them, not $$3$$:

So let’s add $$22$$ to both sides to make that happen!

$$x^{2}-10x+3=0$$

$$x^{2}-10x+25=22$$

$$(x-5)^{2}=22$$

$$x-5=\sqrt22$$    or    $$x-5=-\sqrt22$$

$$x=5+\sqrt 22$$    or    $$x=5- \sqrt 22$$  □

EXAMPLE 35: Solve $$x^{2}-6x=55$$.

Answer: To obtain a perfect square add $$9$$ to both sides.

$$x^{2}-6x+9=64$$

$$(x-3)^{2}=64$$

$$x-3=8$$    or    $$x-3=-8$$

$$x=11$$    or    $$x=-5$$  □

EXAMPLE 36: Solve $$w^{2}+90=22w-31$$.

Answer: Let’s bring all the terms containing a variable to the left side:

$$w^{2}-22w+90=-31$$.

The box tells us to add $$31$$ to each side:

$$w^{2}-22w+121=0$$

$$(w-11)^2=0$$

$$w-11=0$$

$$w=11$$   □

PRACTICE 37: Solve:

a) $$x^{2}+12x-5=40$$

b) $$z^{2}+2z+3=11$$

c) $$x^{2}-40x+300=69$$

d) $$2f^{2}-16f+30=48$$

e) $$x^{2}+100x+2=3$$

PRACTICE 38: Solve $$x^{2}+3x+1=5$$. Don’t be afraid of fractions. You can handle them!

EXAMPLE 39: Solve $$x^{2}+3x+1=5$$.

Answer: Solving this problem does indeed require the use of fractions:

Adding $$1\frac{1}{4}$$ to both sides gives:

$$x^{2}+3x+ \frac{9} {4} = 5 + 1 \frac {1} {4}$$

$$(x+\frac {3} {2})^{2} = 6 \frac {1} {4}$$

$$(x+ \frac {3} {2})^{2} = \frac {25} {4}$$

$$x + \frac {3} {2} = \frac {5} {2}$$    or    $$x + \frac {3} {2} = – \frac {5} {2}$$

$$x = 1$$    or    $$x=-4$$

However, most people would prefer to not work with fractions.

The problem here is that the middle term, the $$3x$$, has an odd number for a coefficient. So … If we don’t like that, let’s change it!

Perhaps the easiest way to create an even number in the middle is to multiply everything through by 2, so $$x^{2}+3x+1=5$$ becomes:

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

COMMENT: Some students might say it would be simpler to add $$x$$ to both sides to then obtain $$x^{2} + 4x + 1 = 5+x$$. But then we have $$x$$’s on both sides of the equation, which is annoying.

Okay … Let’s try the box method on $$2x^{2} + 6x + 2=10$$.

But we have a problem now with the very first piece of the equation: We need two symmetrical terms that multiply together to make $$2x^{2}$$. Most students would suggest $$2x$$ and $$x$$, but this right away ruins the square method we’ve been following in levels 1, 2, and 3.

ALTERNATIVE IDEA: Instead of multiplying through by 2, multiply through by 4.

This again makes the middle term even AND solves the problem with the first term. Let’s see why:

Start with: $$x^{2}+3x+1=5$$.

Multiply through by 4:

$$4x^{2}+12x+4=20$$.

Now apply the box method:

We see that $$4x^{2}$$ comes from $$2x$$ multiplied with $$2x$$, preserving the symmetry.

Notice that the middle term $$12x$$ splits into two equal pieces, $$6x$$ plus $$6x$$, as planned, which means we need the numbers $$3$$ and $$3$$ on the sides ( $$2x$$ times THREE gives $$6x$$).

We also see that the number we need from the box is $$9$$. Adding $$5$$ to both sides of the equation $$4x^{2}+12+9=20$$ yields:

$$4x^{2}+12x+9=25$$.

The box shows that  $$4x^{2}+12x+9$$ is really $$(2x+3)^{2}$$, so we have a level 1 problem:

$$(2x+3)^{2}=25$$

$$2x+3=5$$    or    $$2x+3=-5$$

$$2x=2$$    or    $$2x=-8$$

$$x=1$$    or    $$x=-4$$  □

IF THE MIDDLE TERM IS ODD, MULTIPLYING THROUGH BY $$4$$ IS A CLEVER IDEA!

EXAMPLE 40: Solve $$x^{2}-5x+6=2$$.

Answer: Let’s multiply through by $$4$$:

$$4x^{2}-20x+24=8$$

Adding $$1$$ to both sides gives:

$$4x^{2}-20x+25=9$$

$$(2x-5)^2=9$$

$$2x-5=3$$    or    $$2x-5=-3$$

$$2x=8$$    or    $$2x=2$$

$$x=4$$    or    $$x=1$$  □

EXAMPLE 41: Solve $$x^{2}+x=\frac {3} {4}$$.

Answer: Let’s multiply through by $$4$$:

$$4x^{2}+4x=3$$

Adding $$1$$ to both sides gives:

$$4x^{2}+4x+1=4$$

$$(2x+1)^2=4$$

$$2x+1=2$$    or    $$2x+1=-2$$

$$2x=1$$    or    $$2x=-3$$

$$x=\frac {1} {2}$$  or  $$x=-\frac {3} {2}$$  □

EXAMPLE 42: Solve $$p^{2}+7p-2=5$$.

Answer: Let’s multiply through by $$4$$:

$$4p^{2}+28p-8=20$$

Adding $$57$$ to both sides gives:

$$4p^{2}+28p+49=77$$

$$(2p+7)^2=77$$

$$2p+7=\sqrt 77$$    or    $$2p+7=-\sqrt77$$

$$2p=\sqrt77 – 7$$    or    $$2p=-\sqrt77 – 7$$

$$p=\frac {\sqrt 77 – 7} {2}$$    or    $$p=\frac {-\sqrt 77 – 7} {2}$$  □

PRACTICE 43: Solve:

a) $$x^{2}+11x-5=7$$

b) $$z^{2}-3z+1=-1$$

c) $$x^{2}-x-1=2 \frac {3} {4}$$

d) $$x^{2} +5x+12=70$$

e) $$x^{2}+3=9$$

PRACTICE 44: OPTIONAL CHALLENGE

At the beginning of this level we first tried solving $$x^{2}+3x+1=5$$ by multiplying through by $$2$$ to obtain an “even middle term:” $$2x^{2}+6x+2=10$$.

The trouble with this is that $$2x^{2}$$, at first glance, doesn’t seem to be the product of two symmetrical terms:

However, if we are willing to work with square roots, it is!

The side term must be a number which multiplies by $$\sqrt2 x$$ to give $$3x$$.  That would be $$\frac {3} {\sqrt2}$$.

The box method is so robust that it will work no matter how or when you choose to try it. Continue the box method to show that our equation $$2x^{2}+6x+2=10$$ can be rewritten $$(\sqrt 2 x+\frac {3} {\sqrt 2})^{2}= \frac {25} {2}$$, and that this, by level 2, again has solutions $$x=1$$ and $$x=-4$$.

OKAY .. Now we are ready for the full power of solving quadratics!

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