math help please! Thank you?

Question

equations quadratic form

1. 3(2x - 3)^2 - 5(2x - 3) + 2 = 0

2. x^4 - 9x^2 + 18 = 0

3. Solve the inequality: 4 - x^2 > 0

if you could teach step by step if not okay.

Answer 1

1. let's say X=2x -3 then the equation becomes:
3X^2 - 5X + 2 = 0
divide each side by 3
X^2 - 5/3 X + 2/3 = 0 complete the square what do we need to make X^2 - 5/3X of the form : a^2 - 2ab + b^2 we need to add (5/6)^2 and substract it too
X^2 - 5/3 X + (5/6)^2 - (5/6)^2 + 2/3 = 0
(X - 5/6)^2 - (25/36 - 2/3) = 0
(X - 5/6)^2 - (25/36 - 24/36) = 0
(X - 5/6)^2 - 1/36 = 0
now you have the form A^2 - B^2 which is equal to (A+B)(A-B)
(X - 5/6 + 1/6)(X - 5/6 - 1/6) = 0
(X - 2/3)(X - 1) = 0
so X = 2/3 and 2x - 3 = 2/3 => 2x = 11/3 => x = 11/6
so X = 1 so 2x-3=1 => 2x=4 => x=2

2. x^4 - 9x^2 + 18 = 0
replace x^2 by X
X^2 - 9X + 18 = 0
complete the square X^2 - 9X
X^2 - 9X + (9/2)^2 - (9/2)^2 + 18 = 0
(X - 9/2)^2 - (81/4 - 72/4) = 0
(X - 9/2)^2 - 9/4 = 0
(X - 9/2 - 3/2)(X - 9/2 + 3/2) = 0
(X - 6)(X - 3) = 0
X=6 => x^2 = 6 so x = + or - sqrt6
X = 3 => x^2 = 3 so x = + or - sqrt 3

3. 4 - x^2 > 0
4 > x^2 or x^2 < 4
so x <2 and x>-2 (we know that x^2 = 4 has a double solution for x= + or - 2 so x<2 and x>-2 we change the sign coz of the -)
therefore -2 < x < 2

I hope this helps

Answer 2

The first two equations can be solved with aid of substitution. Let's examine one of a time.

1. 3(2x - 3)^2 - 5(2x - 3) + 2 = 0

Let y = 2x - 3. Then, the equation boils down to 3y^2 - 5y + 2 = 0. Solving for y, we get (solve by yourself!) solutions 1 and 2/3.

Now we have two equations:
y = 1
y = 2/3

Since y = 2x - 3, this is the same as

2x - 3 = 1
2x - 3 = 2/3

Solving each equation, we get x = 2 and x = 11/4. Check the solutions against the original equation, to see if they fit. The one(s) which fit are the solutions to the equation.


2. x^4 - 9x^2 + 18 = 0

In the same fashion, call y = x^2; then, the equation turns to be y^2 - 9y + 18 = 0, whose solutions are (calculate them yourself!) 6 and 3. Then, we have two equations:

y = x^2 = 6
y = x^2 = 3

The solutions for x are, thus, sqrt(6), -sqrt(6), sqrt(3), -sqrt(3); again, check back all of them in the original equation.


3. Solve the inequality: 4 - x^2 > 0

First, solve the corresponding equation: 4 - x^2 = 0, which has solutions -2 and 2. Then, draw a horizontal line, and mark the roots in it: (pardon me the bad ASCII art)

- + -
x: --------------------o--------------------o--------------------
-2 +2

The points divide the line in intervals; in all points of each interval, the value of 4 - x^2 is either positive or negative (the zero values are the points marked). Choose one point x in each interval, put into 4 - x^2, and see the result. The interval(s) where 4 - x^2 > 0 are the solution; in this case, the open interval ]-2, +2[.

Hope this helps.

Answer 3

putting 2x-3=z the equ.becomes

2z^2 -5z +2=0 z=((5+- sqrt(25-16))/4 so z= 2 or z=1/2

2x-3=2 yields x= 5/2 2x-3 =1/2 yields x= 7/4

2) put x^2=z
you get z^2-9z+18 =0

x^2=6 so x= +- sqrt6 x ^2=3 so x= +-sqrt3. You have four values of x

4-x^2 >0 x^2<4 -2
you shoul look at the sign of each factor.At the interval wher both facror have different sign their product or 4x^-2 would be negative

Answer 4

in case you meant 2x/(3 - a million) = 7/x 2x/(3 - a million) = 7/x 2x×x = 7×2 2x^2 = 14 x^2 = 7 x = ±?7 so x = 2.64575 or -2.64575 Or did u mean 2x/3 - a million = 7/x 2x/3 - a million = 7/x x(2x/3 - a million) = 7 2x^2/3 - x = 7 x^2 - 3x/2 = 21/2 (x - 3/4)^2 = 21/2 + 9/sixteen (x - 3/4)^2 = 168/sixteen + 9/sixteen (x - 3/4)^2 = 17716 x - 3/4 = ±?(177/sixteen) x = 3/4 ± ?(177/sixteen) so x = 4.07603 or -2.57603 wish this helps you. *??*

Answer 5

3. Solve the inequality: 4 - x^2 > 0
-x^2>-4
x>2

Is this your home work? I have tried to solve the simplest one; If there are no answers to the more difficult ones, I'll try to provide more answers later.

Answer 6

I don't know Im only in 7th grade.