Honors Algebra II
Solving Equations Using
Quadratic Techniques
Remember that first check to see if the equation can be written
in quadratic form. A helpful hint is to see if you have a trinomial
and in descending degree with the middle terms exponent exactly
half of the leading terms exponent.
Next chose a variable, I like a for this set of problems, the represent
the middle terms variable and exponent.
Then write the equation in quadratic form and solve the quadratic
equation.
DO NOT FORGET TO CHECK YOUR ANSWERS BACK INTO
THE ORIGINAL PROBLEM!!
1. Yes, this equation can be written in quadratic form.
x  13x  36  0
4
2
Let a  x
2
a 2  13a  36  0
a  4a  9  0
a  4 or a  9
ax
2
 4  x2
 4  x
 2i  x
ax
2
9  x
2
 9  x
 3i  x
Because of the corollary to the fundamental theorem of algebra, I
knew that I should have 4 answers, and I did.
2.
x  13x  36  0
4
2
Let a  x
2
a  13a  36  0
a  4a  9  0
a  4 or a  9
2
ax
2
ax
2
4x
2
9x
2
 4x
2 x
 9x
3 x
3.
x 4  50 x 2  49  0
Let a  x
2
a  50a  49  0
a  40a  1  0
a  49 or a  1
2
Did you check your answers?
ax
2
ax
2
49  x
2
1 x
2
 49  x
7  x
 1x
1  x
4.
x  21x  80  0
a  x2
4
Let
2
a  21a  80  0
a  16a  5  0
a  16 or a  5
2
ax
2
ax
2
16  x
2
5 x
2
 16  x
4 x
 5x
5. This one is a little different from the first 4 problems.
2
3
1
3
x  6x  5  0
Let
ax
1
3
a  6a  5  0
2
a  5a  1  0
a  5 or a  1
ax
5 x
1
3
ax
1
3
1


3
5   x 3 
 
125  x
1 x
3
Did you check your answers? How did you check?
1
3
1
3

1   x

1 x
3
1
3




3
6. This one is like the first 4 except that I need to set the equation
= to 0.
x  18  11x
4
2
x  11x  18  0
4
2
Let a  x
2
ax
2
ax
2
2x
2
9x
2
 2x
a 2  11a  18  0
a  2a  9 
a  2 or a  9
Did you check your answers?
 9x
3 x
7. How do I restate the square root of x? Watch, it’s on the wall.
x6 x  7
ax
1
2
x  6x  7  0
Let
ax
1
2
a 2  6a  7  0
a  1a  7   0
a  1 or a  7
1  x
1
2
1
2

2
 1   x

1 x
Did you check your answers?
ax
7x
1
2




2
1
2
1
2

7   x

49  x
2
1
2




2
If you check your answers to # 7 you had to reject x = 1.
How did you check your answers?
If you check your answers to # 7 you had to reject x = 1.
How did you check your answers? I used my handy dandy calc!
What does this graph tell me?
How about this one?
If you check your answers to # 7 you had to reject x = 1.
How did you check your answers? I used my handy dandy calc!
What does this graph tell me?
I need to reject x = 1.
How about this one?
x = 49 is a zero.
8. This one is similar to # 5.
1
2
1
4
x  8 x  15  0
Let
ax
1
4
a  8a  15  0
2
a  5a  3  0
a  5 or a  3
ax
5 x
1
4
1
4

5   x

625  x
4
ax
3 x
1
4




4
Did you check your answer? How did you check?
1
4
1
4

3   x

81  x
4
1
4




4
9. Similar to # 7.
x  8 x  240
x  8 x  240  0
1
2
x  8 x  240  0
Let
ax
1
2
a 2  8a  240  0
a  12a  20  0
a  12 or a  20
ax
 12  x
1
2
ax
1
2

2
 12    x

144  x
20  x
1
2




2
1
2
1
2

20   x

400  x
Check your answers, one of them does not work? Why?
2
1
2




2
10. Similar to # 8.
2
3
1
3
x  11x  28  0
Let
ax
1
3
7  x
a  11a  28  0
2
a  7 a  4  0
a  7 or
ax
a  4
Did you check your answers?
1
3
1
3

 7    x

 343  x
3
ax
4 x
1
3




3
1
3
1
3

 4   x

 64  x
3
1
3




3
11.
x 3  343  0
x  7 x 2  7 x  49  0
x  7
or
x  7 x  49  0
2
7   147
x
2
7  i 49 3
x
2
7  7i 3
x
2
I could have used synthetic division to find the quadratic equation.
How?
# 11 revisted
x 3  343  0
x  343
x  7
3
I know that 7 is a root (or zero). I can now
do synthetic division to get the quadratic equation.
7 1
0
0
343
 7 49  343
1  7 49
0
I get the quadratic equation x2 – 7x + 49 = 0. Now solve
using the quadratic formula, like I did on the previous slide.
12. Similar to # 11.
y 3  512  0
 y  8y 2  8 y  64  0
y  8 or
y  8 y  64  0
2
 8   192
y
2
 8 i 64 3
y

2
2
8i 3
y  4 
2
y  4  4i 3
You could use synthetic
division to find the
quadratic equation if you
didn’t see that this was
the difference of two
cubes.