Algebra help? (5 problems)?

anonymous

1. Solve for x: x^2+6x+9 = 0

2. Solve for x: x^2+2x-8 = 0

3. Solve for V2: (P1V1)/T1 = (P2V2)/T2

4. Solve for L: T = 2π√(L/g)

5. Solve for θ2: n1sinθ1 = n2sinθ2

Please show how you solved them too; I'm trying to learn. Thank you!

m-m

You need to factor for numbers 1 and 2, so if ax^2+bx+c=0 you need to find two numbers that the sum of them equals b and the product of them is c.

1. Since a=1 you must set up your equation as (x )(x )=0 because one times one equals one.

Now you find the numbers I was talking about above. In this case it is 3 and 3.

so: (x+3)(x+3)=0

Anything times zero equals zero, so you must solve what's in each parentheses to equal zero, in this case they are the same so it's just x+3=0, so x=-3

2. a still equals 1 so (x )(x )=0

The two numbers are 4 and -2.

so: (x-2)(x+4)=0

x-2=0 which means x=2

x+4=0 so x can also eqaul -4.

3. To solve for V2 you just need to get it by itself. It is now attached to P2/T2. Anything multiplied by its reciprocal equals one and 1V2 just equals V2. So you multiply both sides by the reciprocal of P2/T2 which is T2/P2.

The final answer is: V2=(P1V1T2)/(T1/P2)

4. First you have to divide both sides by 2π.

T/2π=√(L/g)

Now you have to square both sides to get rid of the square root.

T^2/4π^2=L/g

Now you multiply both sides by the reciprocal of 1/g as you did above, which is just g, to get your final answer.

(gT^2)/4π^2

5. First you divide by n2.

(n1sinθ1)/n2=sinθ2

Then you have to inverse sine both sides (on the calculator it is 2nd sin and looks like sin^-1)

θ2=sin^-1([n1sinθ1]/n2)

allie

With these problems of the form x^2+x+c first you try to factor and if that doesn't work then you use the quadratic equation.

1. So factor it. Set up two bubbles (x+ )(x+ ) now you ask what two numbers add to 6 and multiply to 9. That would be 3 and 3 so your factoring yields (x+3)(x+3)=(x+3)^2 You can check that you've done it right by doing the FOIL method (which stands for First, Outer, Inner, Last) or just simple distribution (which I prefer) If you distribute that first x (ie pretend you just have x(x+3) for a second) you get x^2 + 3x now you distribute the 3 (again pretend you just have 3(x+3)) you get 3x+9. Add both answers up and you get x^2+6x+9 as you should.

2. Same procedure as 1 but now you ask what two numbers add to 2 and multiply to -8. That would be -4 and 2. So your answer is (x - 4)(x+2)

3.Multiply both sides by T2 get (T2P1V1)/T1=(T2P2V2)/T2 the T2 cancels on the right so =P2V2 now divide both sides by P2 get (T2P1V1)/(P2T1)=V2

4.Divide both sides by 2π get T/(2π)=√(L/g) then square both sides to get rid of the root. (T/(2π))^2=T^2/(4π^2)=L/g then multiply both sides by g get gT^2/(4π^2)=L

5.Divide both sides by n1 get n1sin(θ1)/n2=sin(θ2). Trig functions are special and to cancel them you need to do the inverse trig function to both sides. For sin, the inverse trig function is written as arcsin() or sin^-1(). When you do this the argument part (that is the part in the parenthesis) of the sin becomes the whole side. (Like the sin parenthesis eat the whole side of the equation) so you get arcsin((n1sin(θ1))/n2)=arcsin(sin(θ2)) arcsin cancels with sin and pops out the argument so =θ2 You might be able to do some more simplification on the left side, but I can't remember it's been awhile since I was in trig.

Phew

stephen-y

1. Solve for x: x^2+6x+9 = 0

(x+3)(x+3) = 0

x + 3 = 0 ; x = -3

*************************

2.) Solve for x: x^2+2x-8 = 0

(x+4)(x-2) = 0

x + 4 = 0 ; x = -4

x -2 =0 ; x =2

********************************

3,)Solve for V2: (P1V1)/T1 = (P2V2)/T2

P2V2 = (p1V1T2)/T1

V2 =( P1V1T2)/(T1P2)

***************************************************

4. Solve for L: T = 2π√(L/g)

T/2Pi = sq rt (L/G)

(L/G) = (T/2pi)^2

L = {(T/2PI)^2}G

**************************************

5. Solve for θ2: n1sinθ1 = n2sinθ2

sine O2 = (n1 sine O2)/n2

then look up the value of the angle O