Option D: Two irrational solutions
Explanation:
The equation is [tex]17+3 x^{2}=6 x[/tex]
Subtracting 6x from both sides, we have,
[tex]3x^{2} -6x+17=0[/tex]
Solving the equation using quadratic formula,
[tex]x=\frac{6 \pm \sqrt{36-4(3)(17)}}{2(3)}[/tex]
Simplifying the expression, we get,
[tex]\begin{aligned}x &=\frac{6 \pm \sqrt{36-204}}{6} \\&=\frac{6 \pm \sqrt{-168}}{6} \\&=\frac{6 \pm 2 i \sqrt{42}}{6}\end{aligned}[/tex]
Taking out the common terms and simplifying, we have,
[tex]\begin{aligned}x &=\frac{2(3 \pm i \sqrt{42})}{6} \\&=\frac{(3 \pm i \sqrt{42})}{3}\end{aligned}[/tex]
Dividing by 3, we get,
[tex]x=1+i \sqrt{\frac{14}{3}}, x=1-i \sqrt{\frac{14}{3}}[/tex]
Hence, the equation has two irrational solutions.
