Question

Math 163 Test I Sample 1. Solve the equations 4 -3 7 a. \frac{4}{x-2} = \frac{-3}{x+5} + \frac{7}{x^2+3x-10} \implies x=-1 b. -2(x-1)^2 - 5 = -15 \implies x = 1 \pm \sqrt{5} by square root property c. 2x^2 - 8x - 5 = 0 \implies x = 2 \pm \frac{\sqrt{26}}{2} by completing square. d. 9 + \frac{3}{x} - \frac{2}{x^2} = 0 \implies x = \frac{1}{3}, \frac{-2}{3} e. 10x^2 + 6x + 1 = 0 \implies x = \frac{-3 \pm 3i}{10} 2. Perform the operation and simplify. a. \frac{1+4i}{5-12i} = \frac{-43+32i}{169} b. (i^5-i^{-1})^2 = -1 c. (\frac{1}{2} + \frac{\sqrt{3}}{2}i)^2 = \frac{-1 + \sqrt{3}}{2}i 3. Solve the equations. a. \sqrt{2x+3} - \sqrt{x+2} = 2 \implies x = 23, -1 (reject) b. x + 2\sqrt{x} - 3 = 0 \implies x = 1 c. x^2 + 3x + 2 = 0 \implies x = -1, x = -8 d. (x^2 - 1)^2 + (x^2 - 1) - 12 = 0 \implies x = \pm \sqrt{5}i, \pm 2 4. Solve each of the following inequality. Graph each solution and write it in interval notation. a. -3(1-x) < 12 b. -1 \le \frac{3-5x}{3} \le 9 c. (4x-1)^2 > 0 d. |1-4x| < 5 e. |2x-5| > 3 f. -3 \le 5-2|x-2| \le 10 g. 6 - 2|3x-7| < 14

          Math 163 Test I Sample
1. Solve the equations
4
-3
7
a. \frac{4}{x-2} = \frac{-3}{x+5} + \frac{7}{x^2+3x-10} \implies x=-1
b. -2(x-1)^2 - 5 = -15 \implies x = 1 \pm \sqrt{5} by square root property
c. 2x^2 - 8x - 5 = 0 \implies x = 2 \pm \frac{\sqrt{26}}{2} by completing square.
d. 9 + \frac{3}{x} - \frac{2}{x^2} = 0 \implies x = \frac{1}{3}, \frac{-2}{3}
e. 10x^2 + 6x + 1 = 0 \implies x = \frac{-3 \pm 3i}{10}
2. Perform the operation and simplify.
a. \frac{1+4i}{5-12i} = \frac{-43+32i}{169}
b. (i^5-i^{-1})^2 = -1
c. (\frac{1}{2} + \frac{\sqrt{3}}{2}i)^2 = \frac{-1 + \sqrt{3}}{2}i
3. Solve the equations.
a. \sqrt{2x+3} - \sqrt{x+2} = 2 \implies x = 23, -1 (reject)
b. x + 2\sqrt{x} - 3 = 0 \implies x = 1
c. x^2 + 3x + 2 = 0 \implies x = -1, x = -8
d. (x^2 - 1)^2 + (x^2 - 1) - 12 = 0 \implies x = \pm \sqrt{5}i, \pm 2
4. Solve each of the following inequality. Graph each solution and write it in interval notation.
a. -3(1-x) < 12
b. -1 \le \frac{3-5x}{3} \le 9
c. (4x-1)^2 > 0
d. |1-4x| < 5
e. |2x-5| > 3
f. -3 \le 5-2|x-2| \le 10
g. 6 - 2|3x-7| < 14

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Math 163 Test I Sample
1. Solve the equations
4
-3
7
a. (4)/(x-2) = (-3)/(x+5) + (7)/(x^2+3x-10) x=-1
b. -2(x-1)^2 - 5 = -15 x = 1 ±√(5) by square root property
c. 2x^2 - 8x - 5 = 0 x = 2 ±(√(26))/(2) by completing square.
d. 9 + (3)/(x) - (2)/(x^2) = 0 x = (1)/(3), (-2)/(3)
e. 10x^2 + 6x + 1 = 0 x = (-3 ±3i)/(10)
2. Perform the operation and simplify.
a. (1+4i)/(5-12i) = (-43+32i)/(169)
b. (i^5-i^-1)^2 = -1
c. ((1)/(2) + (√(3))/(2)i)^2 = (-1 + √(3))/(2)i
3. Solve the equations.
a. √(2x+3) - √(x+2) = 2 x = 23, -1 (reject)
b. x + 2√(x) - 3 = 0 x = 1
c. x^2 + 3x + 2 = 0 x = -1, x = -8
d. (x^2 - 1)^2 + (x^2 - 1) - 12 = 0 x = ±√(5)i, ±2
4. Solve each of the following inequality. Graph each solution and write it in interval notation.
a. -3(1-x) < 12
b. -1 ≤(3-5x)/(3) ≤9
c. (4x-1)^2 > 0
d. |1-4x| < 5
e. |2x-5| > 3
f. -3 ≤5-2|x-2| ≤10
g. 6 - 2|3x-7| < 14

Added by Megan F.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert Jessica Horn

05/09/2022

Step 1

43 - 7I = 10x - 10 Show more…

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I dont understand whats being asked. The questions given in class are not at all in the same format. I dont get if the answer is already given or what to do. Please help my exam is tomorrow and all my studying doesnt feel like its going to pay off after reviewing this Math 163 Test I Sample 1.Solve the equations 43 7 I-=x x-2x+5x+3x-10 b.-2x-1-5=-15=x=15by square root property 26 c.2x-8x-5=0=x=2 by completing square. 2 1 xx 3. -3i e.10x+6x+1=0=x= 10 2.Perform the operation and simplify 1+4-43+32 3-12 169 b.=-1 1 3.Solve the equations. a.2x+3-x+2=2=x=23.-1(reject b.x+2x-3=0=x=1 g-=x-x=0=2++ dx-1+x-1-12=0=x=3i2 4.Solve each of the following inequality.Graph each solution and write it in interval notation. a.-31-x<12 2 c.4x-1>0 d.1-4x<5 e.2.x-5>3 f.-35-2x-2>10 g6-23x-7<14

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Transcript

00:01 Okay, so here's the deal.

00:03 I thought i was recording and i wasn't, but i really would rather not go back and redo all these.

00:08 So i'm just going to explain to you what i did while i thought i was recording and i wasn't.

00:12 So i started out with the first one, divide on a, dividing everything by two, just to get rid of that coefficient of the two.

00:18 That left me was z squared minus 8z minus 884, equals zero.

00:24 Then i completed the square to get my perfect trunch square here, which gave me 16, which factors to z minus 4 squared.

00:34 And then since i added 16 to one side, i had to add it to the other, and that gave me 900.

00:39 Took the square root of both sides, so z minus 4 equals plus or minus 30.

00:43 So i set it both equal, i did z minus 4 equals 30 and z minus 4 equals negative 30, and i got 34 and 26.

00:52 On the next one, move the 9, move the net 10 over made at negative 10, completed the square, found that to be 9 on both sides, and that factor to x plus 3 squared.

01:02 Negative 10 plus 9 is negative 1.

01:05 Subtrak 3 on both sides.

01:06 So i got negative 3 plus i and negative 3 minus i.

01:10 The next one said use the quadratic formula.

01:13 So i had to put it in standard form first.

01:15 So 3t squared minus 12t plus 15 equals 0.

01:19 So this is my quadratic formula where i plugged in all the pieces.

01:23 And i found t to equal to plus i and two minus.

01:30 I.

01:32 Next, on b, i cross -multiplied.

01:35 When i cross -multiplied, i got 4v minus 16 equals v squared minus 6v, and then i moved everything over to get it equal to zero and had v squared minus 10b plus 16.

01:48 Plug that into the quadratic formula.

01:52 So, and i found it to equal 10 plus or minus 6 over 2, so 10 plus 6 over 2 is 8, and 10 minus 6 over 2.

02:01 Or two is two.

02:03 So these were my answers for v.

02:05 So v equal eight and that and two.

02:08 Now here what i did is i wanted to get rid of this w first.

02:13 So instead of distributing and then going back and factoring all that good jazz, i just took the w and divided by w.

02:20 But when i do that, i have to account for w equals zero.

02:24 So there is going to be a factor of w equals zero...

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