1001 Algebra Problems.PDF

(Marvins-Underground-K-12) #1

  1. c.Isolate the squared expression on one side,
    take the square root of both sides, and solve
    forx:


–3x^2 = –9
x^2 = 3
x= ± 3 


  1. b.Take the square root of both sides, and solve
    forx:


(4x+ 5)x^2 = –49

4 x+ 5 = ±–49= ±7i


4 x= –5 ± 7i

x=


  1. c.Take the square root of both sides, and solve
    forx:


(3x– 8)^2 = 45
3 x– 8 = ± 45 
3 x= 8 ± 45 

x= =


  1. c.Isolate the squared expression on one side,
    take the square root of both sides, and solve
    forx:


(–2x+ 1)^2 – 50 = 0
(–2x+ 1)^2 = 50
–2x+ 1 = ± 50 
–2x= –1 ± 50 
x= = =


  1. b.Isolate the squared expression on one side,
    take the square root of both sides, and solve
    forx:


–(1 – 4x)^2 –121 = 0
(1 – 4x)^2 = –121
1 – 4x= ±–121
–4x= –1 ±–121

x= = =


  1. a.To solve the given equation graphically, let y 1
    = 5x^2 – 24,y 2 = 0. Graph these on the same set
    of axes and identify the points of intersection:


The x-coordinates of the points of intersection
are the solutions of the original equation. We
conclude that the solutions are approximately
±2.191.


  1. d.To solve the equation graphically, let y 1 = 2x^2 ,
    y 2 = –5x– 4. Graph these on the same set of
    axes and identify the points of intersection:


The x-coordinates of the points of intersection
are the solutions of the original equation. Since
the curves do not intersect, the solutions are
imaginary.

y 2 y^1

–5 –4 –3 –2 –1 1 2 3 4 5
–2
–4
–6
–8
–10

10
8
6
4
2

y^1

–4–3–2–1 1 2 3 4 y^2
–3
–6
–9
–12
–15
–18
–21
–24

9
6
3

1 ±11i
4
1 ±–121
4
–1 ±–121
–4

1 ±5^2 
2
1 ±^50 
2
–1 ±^50 
–2

8 ±3^5 
3
8 ±^45 
3

–5 ±7i
4

ANSWERS & EXPLANATIONS–
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