Question

If the equations \[ ax^2+2bx+3c=0,\qquad 3x^2+8x+15=0 \] have a common root, where \(a,b,c\) are sides of triangle \(ABC\), then \[ \cos2A+\cos2B+\cos2C= \]

• A. \(2\)
• B. \(1\)
• C. \(-2\)
• D. \(-1\)

Hint:

If the sides of a triangle are in the ratio \[ \boxed{3:4:5,} \] then the triangle is right-angled. Hence, \[ \boxed{\cos2A+\cos2B+\cos180^\circ=-1.} \]

Answer 1

The Correct Option is D

Step 1: Use the common root condition. Let the common root be \(\alpha\). Since \[ 3\alpha^2+8\alpha+15=0, \] we have \[ \alpha^2=-\frac{8\alpha+15}{3}. \] Substituting into \[ a\alpha^2+2b\alpha+3c=0, \] gives \[ a\left(-\frac{8\alpha+15}{3}\right)+2b\alpha+3c=0. \] Multiplying by \(3\), \[ -8a\alpha-15a+6b\alpha+9c=0. \] Therefore, \[ (6b-8a)\alpha+(9c-15a)=0. \] Since \(\alpha\) is irrational, both coefficients must vanish. Hence, \[ 6b-8a=0, \] \[ 9c-15a=0. \] Thus, \[ \boxed{a:b:c=3:4:5.} \]

Step 2: Identify the triangle. Since \[ 3^2+4^2=5^2, \] triangle \(ABC\) is right-angled. Taking \[ C=90^\circ, \] we have \[ A+B=90^\circ. \]

Step 3: Evaluate the expression. Using \[ \cos2A+\cos2B = \cos2A+\cos(180^\circ-2A) = \cos2A-\cos2A = 0. \] Also, \[ \cos2C = \cos180^\circ = -1. \] Hence, \[ \cos2A+\cos2B+\cos2C = 0+(-1) = \boxed{-1.} \] Thus, \[ \boxed{(D)} \] is the correct answer.