Download Help Examples for w5 3. Rewrite the equation x 2 − 8x − 35 = 0 by

Help Examples for w5
3. Rewrite the equation x2 − 8x − 35 = 0 by completing the square first and then
find the solutions.
Solution. To complete x2 − 8x into a complete square, we need to add (−8/2)2 = 16
to it and the square is (x−4)2 . Since we already have −35, we will need to add both sides
by 51 in order to get 16 on the left side: x2 − 8x − 35 + 51 = 51. So x2 − 8x +√
16 = 51,
2
which is √(x − 4) = 51. Take square root on both sides, we get x − 4 = ± 51. So
x = 4 ± 51.
4–5. Just use the quadratic formula.
6. By completing the square, write the equation x2 + 16x + 114 = 0 as (x + A)2 = B.
Solution. This is just completing square. Subtracting 114 on both sides: x2 + 16x =
−114. We need to add 82 = 64 to x2 + 16x to make a square (which is (x + 8)2 so just
add 64 to both sides: x2 + 16x + 64 = −114 + 64. Thus (x + 8)2 = −50 so in this case
A = 8 and B = −50.
7. The difference of two positive numbers is 3 and the sum of their squares is 117.
Find the numbers.
Solution. Name the smaller number to be x, then the bigger one is x + 3. The sum
of their squares is x2 + (x + 3)2 = 117. This is a quadratic equation after you simplify it
(multiply out (x + 3)2 and combine the like terms) and you should be able so solve it.
8. NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level,
as a function of time is given by h(t) = −4.9t2 + 130t + 283. Assuming that the rocket
will splash down into the ocean, at what time does splashdown occur?
Solution. When the rocket hits the ocean, its height becomes zero. That is, h(t) = 0.
That means −4.9t2 + 130t + 283 = 0. This is again just a quadratic equation. Use the
quadratic formula and your calculator to obtain a numerical answer.
9. Simplify the expression
x2
1
1
− 2
+ 6x + 8 x − 2x − 8
Solution. x2 + 6x + 8 = (x + 2)(x + 4) and x2 − 2x − 8 = (x − 4)(x + 2) so the LCD
is (x + 2)(x + 4)(x − 4) and
1
1
−
x2 + 6x + 8 x2 − 2x − 8
1
1
=
−
(x + 2)(x + 4) (x − 4)(x + 2)
(x − 4)
(x + 4)
=
−
(x + 2)(x + 4)(x − 4) (x − 4)(x + 2)(x + 4)
(x − 4) − (x + 4)
(x + 2)(x + 4)(x − 4)
−8
x−4−x−4
=
.
=
(x + 2)(x + 4)(x − 4)
(x + 2)(x + 4)(x − 4)
=
10. Solve the equation 3x4 − 3x3 − 3x2 = 0.
Solution. There is a common factor x2 that you need to factor out first: 3x4 − 3x3 −
3x2 = x2 (3x2 − 3x − 3) = 0. Let each factor = 0, you get x2 = 0 and 3x2 − 3x − 3 = 0.
x2 = 0 gives you a solution x = 0 and you can solve 3x2 − 3x − 3 = 0 for the other two
solutions with the quadratic formula.
√
11. Solve the equation 10 − x + x = −2.
√
Solution. Remember that you need to isolate the radical term first:
10 − x =
−x − 2. Now square both sides: 10 − x = (−2 − x)2 = 4 + 4x + x2 . Combine the
like terms: x2 + 5x − 6 = 0. So (x + 6)(x − 1) = 0 and x = −6 and x = 1. However
since we squared both sides in the process,√we need to check for any possible extraneous
solutions. If you plug in x = 1, you get 10 − 1 + 1 = −2, or 4 = −2. Which is not
true. So the only solution is x = −6.
12. Solve the equation
(x − 6)
1
q
(x − 1)
1
+ 4(x − 1) 2 = 0
Solution. Keep in mind that (x − 1) 2 is the same as
q
(x − 1) and this is an equation
q
with fraction where (x − 1) is the only denominator. Multiply both sides by it and
the equation should simplify and becomes easy to solve.
13. The radical terms have been isolated to each side by themselves. So just square
both sides. Be sure to check your answers afterwards.
14. My suggestion for most of you is to skip this one. For the brave ones, here is a
completely worked out example.
√
√
We are going to solve for t : t − 138 − t + 190 = 101
√
√
Step 1 is to rewrite as: t − 138 = t + 190 + 101
Hence (here you√need to square both sides and multiply out the right side): t − 138 =
t + 190 + 2 ∗ 101 ∗ t + 190 + 1012
You still have a square root after squaring, which you should leave on the right side
and take everything to the left (subtract t on both sides, subtract 190 and 1012 on both
sides, the divide both sides by 202 (which is 2 ∗ 101)):
√
−52.1238 = t + 190
Squaring again: 2716.8866 = t + 190
So, the only possible root is t = 2526.8866. It is a(n) EXTRANEOUS root. (Fill in
the second blank with REAL or EXTRANEOUS)
15. Just find the LCD and multiply it on both sides (you need to factor the quadratic
denominator before you can find the LCD).
16, 18, 19. These are basic and easy quadratic equations. Pay attention to how
you are asked to enter the answers!
17. The LCD should be easy to find. Multiply it on both sides.
20. Multiply both sides out and combine the like terms.
21. This is similar to #10.
22. Similar to #15.
23. You drop a rock into a deep well. You cant see the rocks impact at the bottom,
but you hear it after 6 seconds. Find the depth of the well is feet. Ignore air resistance.
Solution. The time that passes after you drop the rock has two components: the
time it takes the rock to reach the bottom of the well, and the time that it takes the
sound of the impact to travel back to you.
Let d be the depth of the well (in feet) and let t be the time it takes for the rock to
reach the well water level. The basic physics tells us that d = 16t2 . Since the speed of
sound is 1100 feet per second. The time it takes for the sound to travel back from the
water level to you is d/1100 = 16t2 /1100. So the total time for you to hear the sound
of the rock is t + 16t2 /1100. We are told that we heard the rock in 6 seconds. So the
equation becomes t + 16t2 /1100 = 6. Multiply both sides by 1100 and combine the like
terms: 16t2 + 1100t − 6600 = 0. Solve this quadratic equation with the formula and your
calculator for t first, then get the depth of the well using d = 16t2 (plug in the t value
you solved). In this particular example, I have t = 5.552 so d ≈ 493 feet.
24. Set C(x) = R(x) and solve the quadratic equation using the quadratic formula
and your calculator.