Question

X^2-x=12 -(1/5)x^2+45=0 4/3x^2-48=0

Answer 1

Решение внизу на фото

Answer 2

1)X^2-x=12

x^2-x-12=0

x^2+3x-4x-12=0

x*(x+3)-4(x+3)=0

(x+3)(x-4)=0

x1=-3

x2=4

2)-(1/5)x^2+45=0

-x^2+225=0

-x^2=-225

x1=-15

x2=15

3)4/3x^2-48=0

4x^2-144=0

4x^2=144

x1=-6

x2=6

Answer 3

To solve the given equations, we need to find the values of 'x' that satisfy each equation. Let's solve them one by one:

1. $x^2 - x = 12$

To solve this quadratic equation, we need to set it equal to zero:

$x^2 - x - 12 = 0$

Now, we can factor the equation:

$x^2 - x - 12 = (x - 4)(x + 3) = 0$

Now, set each factor equal to zero and solve for 'x':

1. $x - 4 = 0$ $x = 4$

2. $x + 3 = 0$ $x = -3$

So, the solutions for $x$ in the equation $x^2 - x = 12$ are $x = 4$ and $x = -3$.

2. $-\frac{1}{5}x^2 + 45 = 0$

To solve this quadratic equation, we need to set it equal to zero:

$-\frac{1}{5}x^2 + 45 = 0$

To eliminate the fraction, we can multiply both sides by 5:

$-x^2 + 225 = 0$

Now, let's factor the equation:

$-x^2 + 225 = -(x^2 - 225) = -(x - 15)(x + 15) = 0$

Now, set each factor equal to zero and solve for 'x':

1. $x - 15 = 0$ $x = 15$

2. $x + 15 = 0$ $x = -15$

So, the solutions for $x$ in the equation $-\frac{1}{5}x^2 + 45 = 0$ are $x = 15$ and $x = -15$.

3. $\frac{4}{3}x^2 - 48 = 0$

To solve this quadratic equation, we need to set it equal to zero:

$\frac{4}{3}x^2 - 48 = 0$

To eliminate the fraction, we can multiply both sides by 3:

$4x^2 - 144 = 0$

Now, let's factor the equation:

$4x^2 - 144 = 4(x^2 - 36) = 4(x - 6)(x + 6) = 0$

Now, set each factor equal to zero and solve for 'x':

1. $x - 6 = 0$ $x = 6$

2. $x + 6 = 0$ $x = -6$

So, the solutions for $x$ in the equation $\frac{4}{3}x^2 - 48 = 0$ are $x = 6$ and $x = -6$.

To summarize, the solutions for each equation are as follows:

1. $x^2 - x = 12$ has solutions $x = 4$ and $x = -3$.
2. $-\frac{1}{5}x^2 + 45 = 0$ has solutions $x = 15$ and $x = -15$.
3. $\frac{4}{3}x^2 - 48 = 0$ has solutions $x = 6$ and $x = -6$.