Quadratics

2.4 Level 5 Quadratics – and VICTORY!

Lesson materials located below the video overview.

Here we go!

LEVEL 5 QUADRATICS:
VICTORY OVER QUEEN QUADRIKA AND HER LEGIONS OF SQUAREKS, EXPANDICONS, AND QUADRILITES 

 

EXAMPLE 45: 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.

We have in the previous level multiplied through by \(4\) and have successfully dealt with \(4x^{2}\) as \(2x \times 2x\). This works because \(4\) is a perfect square.

So in this problem, let’s try making \(3x^{2}\) 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 fractions, 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!

Q2_4_pic1

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 46: 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):

\(196x^{2}-28x+28=252\)

Q2_4_pic2

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 47: 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.

Q2_4_pic3

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:

\(16x^{2}+24x-24=40\).

Q2_4_pic4

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 BOX

and off you go!

THE BOX METHOD WILL NEVER LET YOU DOWN!

 

EXAMPLE 48: Solve \(-2x^{2}+3x+7=1\).

 

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

\(16x^{2}-24x-56=-8\)

Q2_4_pic5

\(16x^{2}-24x+9=-8+9+56\)

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

\(4x-3=\pm \sqrt57\)

\(x=\frac {3 \pm \sqrt57} {4}\)

Done!    □   (QUESTION: Was it dangerous to use the \(\pm\) symbol here?)

 

EXAMPLE 49: Solve \(11x^{2}-x+5=0\).

 

Answer: Let’s multiply through by \(11\)and by \(4\), that is, through by \(44\):

\(484x^{2}-44x+220=0\).

Q2_4_pic6

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 thus far 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 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.

 

YOUR TURN …

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

 

PRACTICE 50: 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)  \(4-3x^{2}=2-x\)

 

PRACTICE 51OPTIONAL CHALLENGE

We solved \(3x^{2}=5x+1=9\) in example 45 after rejecting alternative 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!

 

PRACTICE 52: 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?)

 

PRACTICE 53: 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?

 

PRACTICE 54: 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 g).

h) \(3\beta – 2\sqrt\beta=7\)

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

 

PRACTICE 55:

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 \(y\)?

 

 

 

Self Check: Question 6

A chance to check our abilities!

Self Check: Question 7

Another question!

Self Check: Question 8

A third question!

Self Check: Question 9

Yet another question!

Self Check: Question 10

And why not another?

 

 

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