The roots of the equation 7x2 - 6x + 1 = 0 are tanα and tanβ, where 2α and 2β are the angles of a triangle. Which one of the following is correct?
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The triangle is equilateral.
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The triangle is isosceles but not right-angled.
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The triangle is right-angled.
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The triangle is right-angled isosceles.
tan(alpha) + tan(beta) = 6/7, tan(alpha)*tan(beta) = 1/7. tan(alpha+beta) = (tan(alpha)+tan(beta)) / (1 - tan(alpha)*tan(beta)) = (6/7) / (1 - 1/7) = (6/7) / (6/7) = 1. Thus alpha+beta = 45 degrees. Since 2alpha and 2beta are angles of a triangle, 2alpha + 2beta = 90 degrees. The third angle is 180 - 90 = 90 degrees, so it is right-angled.
Using the sum and product of roots formulas for 7x^2 - 6x + 1 = 0, we find that tan(a) + tan(b) = 6/7 and tan(a)tan(b) = 1/7. The double angle tangent identity states tan(2a + 2b) equals (tan(2a) + tan(2b)) divided by (1 - tan(2a)tan(2b)), where tan(2a) = 2tan(a)/(1 - tan^2(a)). We first calculate tan(2a) + tan(2b) = 2(tan(a) + tan(b))/(1 - tan(a)tan(b)) = 2(6/7)/(1 - 1/7) = 2, and we calculate tan(2a)tan(2b) = 4tan(a)tan(b)/(1 + tan(a)tan(b))(1 - tan(a)tan(b)) by using the value to find it equals 1/2. Plugging these into the identity gives tan(2a + 2b) = 2/(1 - 1/2) = 4. Since 2a and 2b are angles of a triangle, the third angle is 180 - 2a - 2b, and tan(180 - x) = -tan(x), meaning the tangent of the third angle is -4. Because the tangent of this third angle is defined and non-zero, the angle must strictly be 90 degrees, making the triangle right-angled.