Download Pre-Calculus Parametrics Worksheet #2

Chapter 2 Review
1. Solve for t: 2t5 + 5t4 – 12 t3 = 0
t3(2t2 + 5t – 12)=0
t3 (2t – 3)(t + 4)=0
t = 0, 3/2, -4
2. Solve for x: √𝑥 2 + 5𝑥 − 2 = 2
square both sides: x2 + 5x – 2 = 4, x2 + 5x – 6=0,
(x+6)(x-1) = 0, x = -6, 1
Check: 36 – 30 – 2 = 4 OK, 1 + 5 – 2 = 4 OK
3. Find the value of k such that the following equation has exactly one real root:
3x2 + (√2𝑘 ) x+ 6 = 0
need b2 – 4ac = 0
2k – 72 = 0, k = 36
4. Show that the quadratic equation ax2 + bx –a = 0, a0, has two distinct real roots.
need b2 – 4ac > 0
b2 + 4a2 > 0; because of squares, left side is >0
5. Find the key numbers and solve the inequality: x4 – 16 < 0
Key points ±2, x < -2 doesn’t work, x > 2 doesn’t work, solution is -2<x<2
6. Solve for x:
√2𝑥 − 1 - √𝑥 − 2 = 1
√2𝑥 − 1 = 1+√𝑥 − 2
2x-1 = 1 + x – 2 + 2√𝑥 − 2 or x = 2√𝑥 − 2
x2 = 4(x-2), x2 – 4x + 8 = 0, x = [4 √16 − 32 ] /2, x = (44i)/2 = 2  2i
Check: √4  4𝑖 − 1 = 1 + √2  2𝑖 − 2 or √3  4𝑖 = 1 + √2𝑖
2
𝑥
7. Solve for x: 𝑥 < 2
4< x2 , key points are x = 0, x = ±2 check: x = -3, 2/(-3) < -3/2, doesn’t work;
x = -1, -2 < -1/2; works; x = 1, 2/1 < ½, doesn’t work, x = 3, 2/3 < 3/2, works
solution is (-2, 0) and 2<x
8. Solve for x:
x3/4 – 6x2 – 2x > 0
x (x2/4 – 6x – 2) >0
(x/4)(x2 – 24x – 8)>0, x= 0, x = [24  √576 + 32]/2 = [24  √604] /2=122√38
x<0, doesn’t hold, 0 < x < 12-2√38 slightly negative, x >0, so holds,
9. Solve for x:
𝑥 2 +1
<0
𝑥 3 −5
3
Numerator > 0, denom for x > √5; 1 key point
3
denom < 0 for x< √5;
10. The sum of the digits in a certain two-digit number is 11. If the order of the digits is reversed, the
number is increased by 27. Find the original number.
a + b = 11, 10a + b = 10b + a + 27
a = 11-b, 110 – 10b + b = 10b + 11 – b + 27
110 – 9b = 9b + 38
110 – 38 = 18b, 72 = 18b, b = 4, a = 7 74 = 47 + 27, yes
ok
11. Solve for x:
1−2𝑥
<½
1+2𝑥
1-2x < ½ + x or ½ < 3x, x > 1/6 or
if ½ + x < 0, x < -1/2
12. Solve for x: √𝑥 2 − 𝑥 − 1 –
2
√𝑥 2 −𝑥−1
=1
let y = x2 – x – 1, √𝑦 – 2/√𝑦 = 1, y – 2 = √𝑦 , (y2 – 4y + 4) = y, y2 – 5y + 4 = 0
(y-4)(y-1) = 0, y = 4, y = 1, x2 – x – 1 = 4, or x2 – x – 5 = 0, x = (1  √1 + 20)/2
13
x2 – x – 1 = 1, x = [1  √1 + 8)]/2 = 2 = 2, −1; check: √𝑥 2 − 𝑥 − 1 = √4 − 2 − 1 = 1,
doesn’t work; √𝑥 2 − 𝑥 − 1 = √1 + 1 − 1 = 1, 1 – 2 = 1; doesn’t work
x = (1 sqrt(21))/2
13. Find a quadratic equation with integer coefficients and the given values as roots:
r1 = (1+√5 )/3, r2 = (1-√5 )/3
r1 r2 = c, -(r1 + r2 ) = b, r1 r2 = (1 – 5 )/9 = -4/9, r1 + r2 = 2/3; x2 – 2x/3 – 4/3 = 0
9x2 – 6x – 4 = 0
14. Solve for x in terms of a and b, a  - b
𝑎
𝑏
𝑥−𝑎
𝑥−𝑏
+
=
+
𝑥−𝑏
𝑥−𝑎
𝑏
𝑎
(xa – a2 + xb – b2)/(x2 – bx – ax + ab) = (ax – a2 + bx – b2)/ab; numerators are the same,
so (x2 – bx – ax + ab) = ab and we have x2 – bx – ax = 0, x(x – a – b) = 0, x = 0 or x = a+b, or
(x(x-b) – a(x-b)) = (x-a)(x-b) = 0,
Alternatively, we could have the numerators be zero: x (a+b) – (a2 + b2) = 0, or x = (a2 + b2)/(a+b)