Question

$|x^2+x+1|+|2x-3|=4x$

Answer 1

{x=1, 2}

Answer 2

The given equation is $|x^2+x+1|+|2x-3|=4x$.

First, consider the term $|x^2+x+1|$. The quadratic expression $x^2+x+1$ has a discriminant $\Delta = b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3$. Since the discriminant is negative and the leading coefficient (1) is positive, the quadratic $x^2+x+1$ is always positive for all real values of $x$. Therefore, $|x^2+x+1| = x^2+x+1$ for all $x \in \mathbb{R}$.

The equation becomes $x^2+x+1+|2x-3|=4x$.

The term $|2x-3|$ depends on the sign of $2x-3$. The critical point is when $2x-3=0$, which means $x=3/2$.

Also, note that the left side of the equation is a sum of absolute values, which must be non-negative. Thus, the right side $4x$ must be non-negative. This implies $4x \ge 0$, so $x \ge 0$. Any solution must satisfy this condition.

We consider two cases based on the sign of $2x-3$, keeping in mind the condition $x \ge 0$. The critical point $x=3/2$ divides the domain $x \ge 0$ into two intervals: $[0, 3/2)$ and $[3/2, \infty)$.

Case 1: $0 \le x < 3/2$.

In this interval, $2x-3 < 0$, so $|2x-3| = -(2x-3) = 3-2x$.

The equation becomes:

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

$x^2-x+4=4x$

$x^2-5x+4=0$

Factoring the quadratic equation:

$(x-1)(x-4)=0$

This gives potential solutions $x=1$ or $x=4$.

We must check if these potential solutions satisfy the condition for this case, which is $0 \le x < 3/2$.

For $x=1$: $0 \le 1 < 3/2$ (True). So $x=1$ is a valid solution in this case.

For $x=4$: $0 \le 4 < 3/2$ (False, since $4 \not< 3/2$). So $x=4$ is not a valid solution in this case.

Case 2: $x \ge 3/2$.

In this interval, $2x-3 \ge 0$, so $|2x-3| = 2x-3$.

The equation becomes:

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

$x^2+3x-2=4x$

$x^2-x-2=0$

Factoring the quadratic equation:

$(x-2)(x+1)=0$

This gives potential solutions $x=2$ or $x=-1$.

We must check if these potential solutions satisfy the condition for this case, which is $x \ge 3/2$.

For $x=2$: $2 \ge 3/2$ (True, since $2 = 4/2$). So $x=2$ is a valid solution in this case.

For $x=-1$: $-1 \ge 3/2$ (False). So $x=-1$ is not a valid solution in this case.

The potential solutions obtained from both cases are $x=1$ and $x=2$. Both of these solutions satisfy the overall condition $x \ge 0$ (since $1 \ge 0$ and $2 \ge 0$).

Finally, we verify the solutions by substituting them back into the original equation:

For $x=1$:

$|1^2+1+1|+|2(1)-3| = |3|+|-1| = 3+1 = 4$.

$4x = 4(1) = 4$.

LHS = RHS, so $x=1$ is a solution.

For $x=2$:

$|2^2+2+1|+|2(2)-3| = |7|+|1| = 7+1 = 8$.

$4x = 4(2) = 8$.

LHS = RHS, so $x=2$ is a solution.

Both $x=1$ and $x=2$ are solutions to the given equation.

Explanation of the solution:

1. Simplify $|x^2+x+1|$ to $x^2+x+1$ as $x^2+x+1 > 0$ for all real $x$.
2. Note that $4x \ge 0$, so $x \ge 0$.
3. Split the problem into cases based on the sign of $2x-3$, considering $x \ge 0$: Case 1 ($0 \le x < 3/2$) and Case 2 ($x \ge 3/2$).
4. Solve the resulting algebraic equations in each case.
5. Check if the solutions obtained in each case satisfy the condition for that case.
6. The valid solutions are $x=1$ (from Case 1) and $x=2$ (from Case 2).
7. Verify these solutions in the original equation. Both $x=1$ and $x=2$ satisfy the equation.