If the equation $ ax^{2} + bx + c = 0 $ and $5x^{2}+ 12x + 13 = 0$ have a common root, where a, b and c are the sides of a triangle $ABC$ then
- $\Delta ABC$ is acute angled
- $\Delta ABC$ is right angled
- $\Delta ABC$ is isosceles
- $\Delta ABC$ is right angled isosceles
If the equations have a common root, the roots of 5x^2 + 12x + 13 = 0 are complex (D = 144 - 260 < 0). If a, b, c are sides of a triangle, they must be real. This implies the equations are identical, so a/5 = b/12 = c/13. Sides are 5k, 12k, 13k. Since 5^2 + 12^2 = 13^2, it is a right-angled triangle.
We find the roots of the equation 5x^2 + 12x + 13 = 0 using the quadratic formula: x = (-12 plus or minus sqrt(144 - 260)) / 10 = (-12 plus or minus sqrt(-116)) / 10. Since the discriminant is negative, the roots are complex non-real numbers. It is mathematically impossible for the real-coefficient equation ax^2 + bx + c = 0 to share a root with this equation unless it shares both complex roots, which requires the two equations to be proportional. Even if they are proportional, the values 5, 12, and 13 form a right-angled triangle because they satisfy the Pythagorean theorem (5^2 + 12^2 = 13^2), giving the result that the triangle is right angled.