Question

X(x-3)=4(x+1)+3x^2-7x

Answer 1

Привет, корней нет.......

Answer 2

Надеюсь помогла xxxxx

Answer 3

To solve the equation $X(x-3)=4(x+1)+3x^2-7x$, let's simplify and find the value(s) of $x$.

Step 1: Expand both sides of the equation: $X(x-3) = X \cdot x - X \cdot 3 = Xx - 3X$

$4(x+1) = 4x + 4$

Step 2: Rewrite the equation with the above expansions: $Xx - 3X = 4x + 4 + 3x^2 - 7x$

Step 3: Collect like terms on one side of the equation: $Xx - 4x - 3X = 3x^2 - 7x + 4$

Step 4: Factor out $x$ from the left side of the equation: $x(X - 4) - 3X = 3x^2 - 7x + 4$

Step 5: Now, let's set the equation to zero and see if we can find the value(s) of $x$: $x(X - 4) - 3X - 3x^2 + 7x - 4 = 0$

Step 6: Combine like terms: $x(X - 3x - 4) + 7x - 3X - 4 = 0$

Step 7: Factor out $X - 4$ from the first term: $x(X - 4) + 7x - 3X - 4 = 0$

Step 8: Now, we want to isolate $x$ on one side of the equation. To do that, let's factor out $x$ from the first two terms and factor out $-1$ from the last two terms: $x(X - 4) + 7(x - 3) = 0$

Step 9: Factor out $x$ from the first two terms: $x(X - 4 + 7(x - 3) = 0$

Step 10: Factor out $X - 4$ from the last two terms: $x(X - 4) + 7(x - 3) = 0$

Step 11: Set each factor to zero and solve for $x$:

$X - 4 = 0 \quad \Rightarrow \quad X = 4$

$x - 3 = 0 \quad \Rightarrow \quad x = 3$

Step 12: Check for extraneous solutions: Substitute $x = 3$ back into the original equation:

$X(x - 3) = 4(x + 1) + 3x^2 - 7x$

$X(3 - 3) = 4(3 + 1) + 3(3)^2 - 7(3)$

$X \cdot 0 = 4 \cdot 4 + 3 \cdot 9 - 21$

$0 = 16 + 27 - 21$

$0 \neq 22$

Since the left-hand side (LHS) and right-hand side (RHS) of the equation are not equal when $x = 3$, this solution is extraneous.

So, the only valid solution is $X = 4$.