Question

1.) 25-16x²=0 2.) 3x²=15 3. )3x²=5 1\3 4.) 4x²=64

Answer 1

25-16x²=0

-16x²=-25

x²=25/16

x=+-√25/16

x=+-5/4

2)3x²=15

x²=15:3

x²=5

x=+-√5

4)4x²=64

x²=64:4

x²=16

x=+-√16

x=+-4

Answer 2

Let's solve each of the quadratic equations one by one:

1.) 25 - 16x² = 0

To solve this equation, we need to move all the terms to one side of the equation and then factorize it:

16x² = 25

Now, divide both sides by 16:

x² = 25/16

To solve for x, take the square root of both sides (remembering to consider both the positive and negative roots):

x = ±√(25/16) x = ±(5/4)

So, the solutions are x = 5/4 and x = -5/4.

2.) 3x² = 15

Divide both sides by 3:

x² = 5

Take the square root of both sides:

x = ±√5

So, the solutions are x = √5 and x = -√5.

3.) 3x² = 5 1/3

First, express the fractional constant on the right-hand side as an improper fraction:

5 1/3 = 5 + 1/3 = 15/3 + 1/3 = 16/3

Now, divide both sides by 3:

x² = 16/3

Take the square root of both sides:

x = ±√(16/3)

To simplify the square root, we can factor 16 as 4 * 4:

x = ±(4/√3)

To rationalize the denominator, multiply both the numerator and denominator by √3:

x = ±(4√3 / 3)

So, the solutions are x = 4√3 / 3 and x = -4√3 / 3.

4.) 4x² = 64

Divide both sides by 4:

x² = 16

Take the square root of both sides:

x = ±√16 x = ±4

So, the solutions are x = 4 and x = -4.