Quadratics

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 \(2x\).

Q2_3_pic1

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 \(4\).

Q2_3_pic2

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 \(=10x\).

Q2_3_pic3

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

Q2_3_pic4

\(x^{2}-10x+25=169\)

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

\(x-5=13\)    or    \(x-5=-13\)

\(x=18\)   or   \(x=-8\)   □

 

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

 

Answer: We have:

Q2_3_pic5

\(x^{2}-20x+100=7\)

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

\(x-10=\sqrt 7\)    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?

Q2_3_pic6

\(x^{2} +2 \sqrt 5 x + 5 = 4\)

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

\(x+\sqrt5 = 2\)    or    \(x+\sqrt5 = -2\)

\(x=2-\sqrt5\)    or    \(x=-2-\sqrt5\)  □

 

YOUR TURN …

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 \(D\).

 

 

LEVEL 3 QUADRATICS

Planet Quadricus!

 

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:

Q2_3_pic7

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.

Q2_3_pic8

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 \(11\):

Q2_3_pic9

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 \(3\):

Q2_3_pic10

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.

Q2_3_pic11

\(x^{2}-6x+9=64\)

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

\(x-3=8\)    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:

\(w^{2}-22w+90=-31\).

Q2_3_pic12

The box tells us to add \(31\) to each side:

\(w^{2}-22w+121=0\)

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

\(w-11=0\)

\(w=11\)   □

YOUR TURN …

 

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!

 

LEVEL 4 QUADRATICS

Fortress Quadrus

 

EXAMPLE 39: Solve \(x^{2}+3x+1=5\).

 

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

Q2_3_pic13

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 \(2x^{2} + 6x + 2=10\).

Q2_3_pic14

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:

Q2_3_pic15

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 \(4\):

\(4x^{2}-20x+24=8\)

Q2_3_pic16

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 \(4\):

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

Q2_3_pic17

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 \(4\):

\(4p^{2}+28p-8=20\)

Q2_3_pic18

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    \(p=\frac {-\sqrt 77 – 7} {2} \)  □

YOUR TURN …

 

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:

Q2_3_pic19

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

Q2_3_pic20

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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