solve for x?

Question

1) 9x^2+4=0
2) x^2/2=x-2

how do i do these..i knw i have to use the quadratic formula but im not sure wat to do if the is no "b" value..and like in number 2) the x^2 is over the 2....can anyone plz help that wold be great...thanx for any help in advance

Answer 1

If there is no b value, just use b = 0.

In the first problem, however, you can solve it without using the quadratic formula, if you like...

PROBLEM 1:

9x² + 4 = 0
9x² = -4
x² = -4/9
x = sqrt(-4/9)
x = sqrt(-4) / sqrt(9)
x = ±2i / 3
x = ±(2/3)i

PROBLEM 2:

x² / 2 = x - 2
x² = 2x - 5
x² - 2x + 4 = 0

a = 1, b = -2, c = 4

...... -(-2) ± sqrt( (-2)² - 4(1)(4) )
x = --------------------------------------
........................ 2(1)

...... 4 ± sqrt( 4 - 16 )
x = ----------------------
................ 2

...... 4 ± sqrt( -12 )
x = ------------------
................ 2

...... 4 ± (2√3)i
x = ---------------
.......... 2

x = 2 ± (√3)i

Answer 2

There is a b value. The b value is zero.

You do not have to use the quadratic formula on the first problem.

1.) 9X² = -4

X² = -4/9

X = ±2/3i

2.) X² = 2X - 4

X² - 2X + 4 = 0

(2 ± √4 - 16) / 2

(2 ± 2i√3) / 2

1 ± i√3

I hope this helped.

Answer 3

9x^2 + 4 = 0
divide by 9 will give you
x^2 +4/9 = 0
x^2 = -4/9
x = +,- square root of - 4/9, so you'll be in the imaginaries

x^2/ 2 = x - 2
bring the x to the left
x^2/2 - x = -2
bring the 2 to the left
x^2/2 - x +2 = 0
multiply by 2
x^2 - 2x + 4 = 0
quadratic formula
-b +- square root(b^2 - 4 ac)/ 2a
2+- sqrt(4- 16)/ 2
2 +- sqrt(-12) / 2
again imaginary

Answer 4

1) not possible, because that would make 9x^2 a negative number. Squares cannot be negative.

2) x^2-2x+4=0 use quadratic formula

Answer 5

The first has no real solution.

1) 9x^2 = -4

Take the square root of either side

3x = 2i -3x = 2i

Where i = sqrt(-1)

So x can be either 2/3*i or -2/3*i

For the second one....

2) x^2 = 2x - 4

x^2 - 2x + 4 = 0

Factor the polynomial above

(x-2)(x-2) = 0

So x = 2 for the second problem.

EDIT:

For the first problem, it IS possible. you just get imaginary results. You get no real roots. But it is possible

Answer 6

for the first one the b value is 0 so work it from there
a = 9
b = 0
c = 4

for the second one I would multiply the entire thing by 2 to get rid of hte fraction
x^2 = 2x - 4

Then move everything over

x^2 - 2x + 4 = 0

Then use the quadratic
a = 1
b = -2
c = 4

Answer 7

If there is no B value, then you substitute 0 for B, it just makes life easier, actually.

If there is x^2 over 2, then it means (x^2) over 2, so you can look at it as if you are multiplying by 1 over 2, or 0.5.

Answer 8

1) a=9, b=0, c=4

2) rearrange x^2/2-x+2=0
a=1/2(or .5), b=-1, c=+2

Answer 9

1) 9x^2+4=0

9x^2 = -4
x^2 = -4/9
x = ...

can't take the square root of negative, unless it is imaginary...
+/- 2/3 * i



2) x^2 = 2(x-2)
x^2 = 2x - 4
x^2 - 2x + 4 =0
x = [2 +/- sqrt(4 - 16) ]/2
x = [ 2+/- sqrt(-12)] / 2
x = 1 +/- i*sqrt(3)

Answer 10

1. no solution
2. cross multiply
2(x-2)=x^2
2x-4=x^2
0=x^2-2x+4
quadratic formula it and u should get the answer