Question

A) 2x2 - 8 = 0; 6) 3x2 - 75 = 0; B) 48 - 3x2 = 0; r) 9x2- 81 = 0. СРОЧНООО ПЖ

Answer 1

Пошаговое объяснение:

a) 2x^2-8=0

2X^2=8

X^2=4

X1= - 2

X2=2

б) 3х^2-75=0

3х^2=75

Х^2=75:3

Х^2=25

Х1= - 5

Х2=5

в)

48-3х^2=0

-3х^2= - 48

Х^2=16

Х1= - 4

Х2=4

г) 9х^2-81=0

9х^2=81

Х^2=81:9

Х^2=9

Х1= - 3

Х2=3

Answer 2

Let's solve each quadratic equation:

A) $2x^2 - 8 = 0$:

1. Add 8 to both sides: $2x^2 = 8$
2. Divide both sides by 2: $x^2 = 4$
3. Take the square root of both sides: $x = \pm 2$

So, the solutions for A) are $x = 2$ and $x = -2$.

6. $3x^2 - 75 = 0$:

1. Add 75 to both sides: $3x^2 = 75$
2. Divide both sides by 3: $x^2 = 25$
3. Take the square root of both sides: $x = \pm 5$

So, the solutions for 6) are $x = 5$ and $x = -5$.

B) $48 - 3x^2 = 0$:

1. Subtract 48 from both sides: $-3x^2 = -48$
2. Divide both sides by -3: $x^2 = 16$
3. Take the square root of both sides: $x = \pm 4$

So, the solutions for B) are $x = 4$ and $x = -4$.

r) $9x^2 - 81 = 0$:

1. Add 81 to both sides: $9x^2 = 81$
2. Divide both sides by 9: $x^2 = 9$
3. Take the square root of both sides: $x = \pm 3$

So, the solutions for r) are $x = 3$ and $x = -3$.

In summary:

• For equation A): $x = 2$ and $x = -2$.
• For equation 6): $x = 5$ and $x = -5$.
• For equation B): $x = 4$ and $x = -4$.
• For equation r): $x = 3$ and $x = -3$.