needs help with algebra 2 hw?

Question

27. 2/*7*-*3* *-square root of
29. (3-2i)/(3+2i)
31. solve by completing the square: 4x^2-12x+7=0
32. Solve by using the quadratic formula: 9y^2+12y+5=0
33. Find the real values of k for which kx^2-6x+3=0 has two imaginary roots.

sorry it's a lot of problems to solve.. please solve the ones you can and include the work process too. thank you :)

Answer 1

27. This may not be correct, because I'm not sure what you mean by the way it is typed.
2/sqrt(7) - sqrt(3)
We need to rationalize the first part, 2/sqrt(7)
Multiply the top and bottom by sqrt(7)
= 2sqrt(7)/7
Answer: 2sqrt(7)/7 - sqrt(3)

29. (3-2i)/(3+2i)
Multiply the top and bottom by the conjugate of the denominator.
(3-2i)(3-2i) / [(3+2i)(3-2i)]
Use FOIL to multiply out the top and bottom.
(9 -6i -6i +4i^2) / (9 -6i +6i -4i^2)
= (9 - 12i -4) / (9+4)
= (5-12i)/13

31. 4x^2 - 12x + 7 = 0
Subtract 7 from each side.
4x^2 - 12x = -7
Divide both sides by 4
x^2 - 3x = -7/4

To complete the square, you divide the coefficient of x by 2, square it, then add that number to each side.
x^2 - 3x + (-3/2)^2 = -7/4 + (-3/2)^2
x^2 - 3x + 9/4 = -7/4 + 9/4
(x -3/2)^2 = 2/4
(x- 3/2)^2 = 1/2
Take the square root of each side.
x - 3/2 = -sqrt(1/2) or x - 3/2 = sqrt(1/2)
Add 3/2 to each side.
x = -sqrt(1/2) + 3/2 or x = sqrt(1/2) + 3/2
Rationalize sqrt(1/2) by multiplying by sqrt(2)/sqrt(2)
x = -sqrt(2)/2 + 3/2 or x = sqrt(2)/2 + 3/2
x = (3 - sqrt(2)/2 or x = (3+sqrt(2)/2

32. 9y^2 +12y +5 = 0
The quadratic formula says that the solutions of
ax^2+bx+c = 0 are x = [-b +/- sqrt(b^2 -4ac)] / (2a)
a = 9, b = 12, c = 5; plug in and simplify.

33. kx^2 -6x +3 = 0
A quadratic equation has imaginary roots if the discriminant, b^2 -4ac < 0.
a = k, b = -6, and c = 3
b^2 - 4ac
= (-6)^2 - 4(k)(3)
= 36 - 12k

We want this to be less than 0.
36 - 12k < 0
Subtract 36 from each side.
-12k < -36
Divide both sides by -12
k > 3

Answer 2

You didn't write 27 clearly.

If the goal is to simplify 29, multiple denominator and numerator by the conjugate of the denominator, which happens to be 3-2i.

For 31, x^2 - 3x + 9/4 is a perfect square, hence so is 4x^2 - 12x + 9. You can proceed from there.

32. Come on. You know the quadratic formula. Just plug in a = 9, b = 12, c = 5, and simplify.

For 33, take the discriminant and see where it is less than 0.

Answer 3

you may first make the sq. root 5 into 5^(a million/2) so which you get .... 5^(a million/2)=5^x then in view which you have a complication-free base (5), you get .... a million/2=x that's by fact if the 5 on the left is raised to the a million/2 potential and that's comparable to the 5 on the left, x could additionally be a million/2 desire this helps :)