Question

Решите уравнение (3x2-8x+5)(2x-3)=0

Answer 1

Відповідь: -5/3; -1; 3/2

Покрокове пояснення:

3х^2-8х+5=0

2х-3=0

1) Д=64-4*3*5=64-60=4

√Д=2

х1=(-8+2)/6=-1

х2=(-8-2)/6=-10/6=-5/3

2)2х=3

х=3/2

Answer 2

To solve the equation $(3x^2 - 8x + 5)(2x - 3) = 0$, we need to find the values of $x$ that make the entire expression equal to zero.

We can do this by setting each factor equal to zero and solving for $x$.

Setting $3x^2 - 8x + 5$ equal to zero:

$3x^2 - 8x + 5 = 0$

This quadratic equation can be factored as:

$(3x - 5)(x - 1) = 0$

Now, setting $2x - 3$ equal to zero:

$2x - 3 = 0$

Now, solve for $x$:

$2x = 3$

$x = \frac{3}{2}$

So, the two solutions to the equation are:

$x = \frac{3}{2}$

and from the factored quadratic:

$3x - 5 = 0 \Rightarrow 3x = 5 \Rightarrow x = \frac{5}{3}$

$x - 1 = 0 \Rightarrow x = 1$

Therefore, the solutions are $x = \frac{3}{2}$, $x = \frac{5}{3}$, and $x = 1$.

Answer 3

To solve the equation $(3x^2 - 8x + 5)(2x - 3) = 0$, we need to find the values of $x$ that satisfy this equation.

The equation is already factored, so we can set each factor equal to zero and solve for $x$.

Setting $3x^2 - 8x + 5 = 0$:

To solve this quadratic equation, we can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $ax^2 + bx + c = 0$.

In our case, $a = 3$, $b = -8$, and $c = 5$.

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3}$

$x = \frac{8 \pm \sqrt{64 - 60}}{6}$

$x = \frac{8 \pm \sqrt{4}}{6}$

$x = \frac{8 \pm 2}{6}$

So, we have two potential solutions:

1. $x = \frac{8 + 2}{6} = \frac{10}{6} = \frac{5}{3}$

2. $x = \frac{8 - 2}{6} = \frac{6}{6} = 1$

Now, setting $2x - 3 = 0$:

$2x = 3$

$x = \frac{3}{2}$

So, the three solutions to the equation are:

1. $x = \frac{5}{3}$
2. $x = 1$
3. $x = \frac{3}{2}$