Question

A)5x+3x³=0б)7(x-3=6(x-2)​

Answer 1

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Answer 2

It looks like you have two equations here:

A) $5x + 3x^3 = 0$ B) $7(x-3) = 6(x-2)$

Let's solve each of them:

A) $5x + 3x^3 = 0$

To solve this equation, we can factor out the common term $x$: $x(5 + 3x^2) = 0$

Now, we have two factors:

1. $x = 0$
2. $5 + 3x^2 = 0$

For the second factor, we can subtract 5 from both sides and then divide by 3 to isolate $x^2$: $3x^2 = -5$

Now, take the square root of both sides, but remember to consider both positive and negative square roots: $x^2 = -\frac{5}{3}$

Taking the square root of both sides: $x = \pm \sqrt{-\frac{5}{3}}$

Since we have a negative value under the square root, this equation has complex solutions. The solutions are:

$x = \sqrt{\frac{5}{3}}i$ (where $i$ is the imaginary unit) $x = -\sqrt{\frac{5}{3}}i$

B) $7(x-3) = 6(x-2)$

Let's distribute the constants on both sides of the equation:

$7x - 21 = 6x - 12$

Now, let's isolate the variable $x$. Start by moving all terms involving $x$ to one side by subtracting $6x$ from both sides:

$7x - 6x - 21 = -12$

Simplify:

$x - 21 = -12$

Next, add 21 to both sides to isolate $x$:

$x = -12 + 21$

$x = 9$

So, the solution to this equation is $x = 9$.