1001 Algebra Problems.PDF

(Marvins-Underground-K-12) #1

  1. c.First, put the equation into standard form
    by expanding the expression on the left side,
    and then moving all terms to the left side of
    the equation:


(3x– 8)^2 = 45
9 x^2 – 48x+ 64 = 45
9 x^2 – 48x+ 19 = 0

Now, apply the quadratic formula with a= 9,
b= –48, and c= 19 to obtain:

x= = =

= =


  1. d.We first multiply both sides of the equation
    by 100, then divide both sides by 20 in order
    to make the coefficients integers; this will help
    with the simplification process. Doing so
    yields the equivalent equation x^2 – 11x+ 10 =

  2. Now, apply the quadratic formula with a= 1,
    b= –11, and c= 10 to obtain:


x= = =

= = 1, 10


  1. d.Apply the quadratic formula with a= 1,
    b= –3, and c= –3 to obtain:


x= = =


  1. b.The simplification process will be easier if
    we first eliminate the fractions by multiplying
    both sides of the equation  61 x^2 – ^53 x+ 1 = 0 by 6.
    Doing so yields the equivalent equation x^2 – 10x


+ 6 = 0. Now, apply the quadratic formula
with a= 1,b= –10, and c= 6 to obtain:

x= = =

= = 5 ± 19 


  1. b.First, put the equation into standard form
    by expanding the expression on the left side,
    and then moving all terms to the left side of
    the equation:


(x– 3)(2x+ 1) =x(x– 4)
2 x^2 – 6x+x– 3 =x^2 – 4x
x^2 – x– 3 = 0

Now, apply the quadratic formula with a= 1,
b= –1, and c= –3 to obtain:

x= = =

Set 41 (Page 97)


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


4 x^2 = 3

x^2 = ^34 

x= ± ^34 = ±^34 = ± 2 ^3


1 ±^13 
2

–(–1) ±(–1)^2 – 4(1)(–3)
2(1)
–b±b^2 –4ac
2 a

10 ± 2^19 
2
10 ±^76 
2

–(–10) ±(–10)^2 –4(1)(6)
2(1)
–b±b^2 –4ac
2 a

3±^21 
2

–(–3) ±(–3)^2 – 4(1)(3)
2(1)
–b±b^2 –4ac
2 a

11 ±9
2
11 ±^81 
2

–(–11) ±(–11)^2 – 4(1)(–10)
2(1)
–b±b^2 –4ac
2 a

–8 ±3^5 
3
–48 ±18^5 
18
–48 ±^1620 
18

–(–48) ±(–48)^2 –4(9)(19)
2(9)
–b±b^2 –4ac
2 a

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