Question

1)3x²-4x=02)4x²-9=03)-5x²+6x=04)-x²+3=0

Answer 1

1)x(3x-4)=0
x=0 3x-4=0
3x=4
x=3÷4
2)?
3)?
4)?

Answer 2

Let's solve each of the quadratic equations you provided:

1. 3x² - 4x = 0

To solve this equation, we can factor out the common term "x" from the left side:

x(3x - 4) = 0

Now we have two factors: x = 0 and 3x - 4 = 0.

Setting 3x - 4 = 0 and solving for x:

3x = 4 x = 4/3

Therefore, the solutions to the equation are x = 0 and x = 4/3.

2. 4x² - 9 = 0

To solve this equation, we can factor it using the difference of squares:

(2x - 3)(2x + 3) = 0

Setting each factor to zero and solving for x:

2x - 3 = 0 2x = 3 x = 3/2

2x + 3 = 0 2x = -3 x = -3/2

Therefore, the solutions to the equation are x = 3/2 and x = -3/2.

3. -5x² + 6x = 0

We can factor out the common term "x" from the left side:

x(-5x + 6) = 0

Now we have two factors: x = 0 and -5x + 6 = 0.

Setting -5x + 6 = 0 and solving for x:

-5x = -6 x = -6/-5 x = 6/5

Therefore, the solutions to the equation are x = 0 and x = 6/5.

4. -x² + 3 = 0

To solve this equation, we can rearrange it to isolate the squared term:

x² = 3

Taking the square root of both sides, considering both positive and negative roots:

x = ±√3

Therefore, the solutions to the equation are x = √3 and x = -√3.