Solve the equation x^2 + 4 times the absolute value of (x - 3), minus 7x, plus 11, equals 0.

Step-by-step solution

1. 1

   Split the line at the point where the modulus changes. $|x-3|$ equals $x-3$ when $x\ge 3$ and $3-x$ when $x<3$. Each case gives a different quadratic, and each quadratic is only valid on its own interval.

2. 2

   Case 1: x >= 3. Substituting $|x-3|=x-3$:

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

3. 3

   Solve Case 1 and test against x >= 3.

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

   Numerically these are $3.3028$ and $-0.3028$. Only the first satisfies $x\ge 3$, so

   $x=\frac{3+\sqrt{13}}{2}\ \text{is accepted, and }\frac{3-\sqrt{13}}{2}\ \text{is rejected.}$

4. 4

   Case 2: x < 3. Now $|x-3|=3-x$:

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

5. 5

   Solve Case 2 and test against x < 3.

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

   numerically $8.1926$ and $2.8074$. Only the smaller one satisfies $x<3$, so

   $x=\frac{11-\sqrt{29}}{2}\ \text{is accepted.}$

6. 6

   Collect and verify the two solutions. Substituting $x=3.30278$: $10.9083+4(0.30278)-23.1194+11=0.0000$ ✓. Substituting $x=2.80742$: $7.8816+4(0.19258)-19.6519+11=0.0000$ ✓. Testing each candidate against the interval that generated it is the step that separates this from an ordinary quadratic — two of the four candidates are extraneous.