Expert: Lynn Houston - 7/19/2009

Question
Hello Lynn,

I have a 350 problem review packet and I have completed most of it but there are several questions I could not come up with answers for.  Many of them are supposed to have the “imaginary i” or square roots in the answer and I’m not getting that. I’m really short on time and I must get this done. :( Can you please help with any of them?

(Note: ^ means to the power of, / means "over" like a fraction bar, and "i" stands for imaginary number.)

Solve: (t-3) ^2=8

Solve: (x-5)^2/3 = -8

Solve: y^2 + 8y + 6= 0

Solve: t^2+8= 4t

Solve: 2n^2-8n -3 =0

Solve: 0.5 x ^2 -3x=2

Solve: 5x(1-x)=5(x-2)

Solve: square root (x+3) = 2x

Simplify: 15y^2 / 15y^3-10y^2

Simplify: a^2-1 / a^2 + 2a+ 1

Simplify:  (6z ^2)^2 / (4z^3)^3

Simplify: (8x^4)^2 / (2x^2)^5

Solve:  5 = 4r (2r+3)

Solve: t^2 + 2= 2t/ 5

Solve: 2w^2 + 4w = -3

Solve: u^2 + 2u-3=0

Solve: z^2+ 2i z-1=0

Solve: 4y^2 + 12y + 9=0

Solve: 5(x+7)^2 = 25

Solve: 10= 6t-t^2

Solve: x^2 + 4x-396 =0

Solve: (2x+5) (x-3) =6

Solve: 2w^2=3(w-2) ^2

Solve: x+1/x – x/ x+ 1=2
Solve: t/ t-2 + 2t/ t-1=6

Find the value of k for which the equation has a real double root:
3x^2-6x-k=0

Find the value of k for which the equation has a real double root:
9x^2-6x+ k=0

Find the value of k for which the equation has a real double root:
3x^2-6kx+12=0

Solve: x^2 - 2x - 5 = 0

Solve: p^2+ 20p+ 200=0

Solve: x/x-1 –x/ x+1=3+2x^2/1-x^2

Solve: 5x^2- 45=0

Solve: 3y^2-48 =0  

Answer
I'll explain some so that you understand the process and then get the rest.  Just eyeballing these, it looks like most of them require the quadratic formula to solve.

(-b +- sq rt (b^2 - 4ac))/2a

where a, b & c are taken from the equation in the form
ax^2 + bx + c = 0


Solve: (t-3) ^2=8  start by doing the math (ie square t-3)
t^2 - 6t + 9 -8 = 0
t^2-6t+1 = 0

now use the values of a, b & c for the quadratic formula

(-(-6) +- sq rt ((-6)^2 - 4(1)(1)))/2(1)
(6 +- sq rt (36 -4))/2

3+ ((sq rt 32)/2) or 3 - ((sq rt 32)/2)
to go further, you need a calculator

x = 5.828 or .172

The next 6 are all solved the same way.  Do all the multiplication so the equation is in the form ax^2 + bx + c = 0 and then use the quadratic formula to solve for x.


Simplify: 15y^2 / 15y^3-10y^2
15y^2 / 5y^2(3y-2)
3/(3y-2)

Simplify: a^2-1 / a^2 + 2a+ 1
(a-1)(a+1) / (a+1)(a+1)
(a-1)/(a+1)

With all the "solve"s under the "simplify"s, you're back to the quadratic eq.

Solve:  5 = 4r (2r+3) (do the math)
5 = 8r^2 + 12r
8r^2 + 12r - 5 = 0 (use the quad formula)
-12 +- sqrt(144+160)/16
-12 +- sqrt304/16
-12 + sqrt304/16 or -12 - sqrt304/16
x = .339 or x = - 1.84

(-b +- sq rt (b^2 - 4ac))/2a

Solve: t^2 + 2= 2t/ 5
t^2 - 2t/5 + 2 = 0 (mult everything by 5)
5t^2 - 2t + 10
2 +- sqrt(4-40)/16 (this is where i comes in)
x=2 + sqrt(-36)/16 or x= 2-sqrt(-36)/16
x= 2 + 6i/16 or x= 2- 6i/16


Find the value of k for which the equation has a real double root:
3x^2-6x-k=0
(3x + 3)(x - 3)
k = 9

I'm sure there's a better way to do these, but I just look at them and since there is only one choice for factors of 3, you know you'll have (3x + a)(x - b).  I started with b=1, then 2, then 3.

Hope these help.