Solve the equation x^2 + 4|x - 3| - 7x + 11 = 0.

Step-by-step solution

1. 1

   Find where the expression inside the bars changes sign. $|x-3|$ behaves differently on either side of $x=3$, so the real line splits into two branches: $x\ge 3$ and $x<3$. On each branch the absolute value can be replaced by an ordinary linear expression, turning one hard equation into two ordinary quadratics.

2. 2

   Branch 1: $x\ge 3$, so $|x-3|=x-3$. Substituting and collecting terms:

   $x^2+4(x-3)-7x+11=x^2+4x-12-7x+11=x^2-3x-1=0$

   $x=\frac{3\pm\sqrt{9+4}}{2}=\frac{3\pm\sqrt{13}}{2}$

3. 3

   Keep only the branch-1 root that actually satisfies $x\ge 3$. Numerically $\frac{3+\sqrt{13}}{2}\approx 3.303\ \ge 3$ (accept), while $\frac{3-\sqrt{13}}{2}\approx-0.303<3$ (reject). The rejected value solves the rewritten equation but not the original one, because on that side $|x-3|$ is not $x-3$.

4. 4

   Branch 2: $x<3$, so $|x-3|=3-x$.

   $x^2+4(3-x)-7x+11=x^2-4x+12-7x+11=x^2-11x+23=0$

   $x=\frac{11\pm\sqrt{121-92}}{2}=\frac{11\pm\sqrt{29}}{2}$

   Here $\frac{11-\sqrt{29}}{2}\approx 2.807<3$ (accept) and $\frac{11+\sqrt{29}}{2}\approx 8.193\ \ge 3$ (reject).

5. 5

   Verify both survivors in the original equation. For $x=\frac{3+\sqrt{13}}{2}\approx3.30278$: $x^2\approx10.9083$, $4|x-3|\approx1.2111$, $-7x\approx-23.1194$, so the total is $10.9083+1.2111-23.1194+11\approx 0$. For $x=\frac{11-\sqrt{29}}{2}\approx2.80742$: $x^2\approx7.8816$, $4|x-3|\approx0.7703$, $-7x\approx-19.6519$, total $\approx 0$. Both check out.

6. 6

   Report both roots. The equation has exactly two solutions,

   $x=\frac{11-\sqrt{29}}{2}\approx2.807\qquad\text{and}\qquad x=\frac{3+\sqrt{13}}{2}\approx3.303$

   One root sits on each side of the break point $x=3$ — a useful reminder that a case split can, and often does, contribute a root from every branch.