Question

Let a,b,c be the three sides of a triangle, then the quadratic equation b²x² +(b² + c² – a²) x + c² = 0 has

• A. Both roots positive
• B. Both roots negative
• C. Both roots imaginary
• D. None of these

Answer 1

Answer: (C)

Both roots imaginary

Answer 2

Q b² + c² – a² = 2bc cosA

\ b²x² + 2bc cosA x + c² = 0

\ D = (2bc cosA)² – 4b²c²

= 4b²c² (cos²A–1) < 0

$x ^ { 2 }$

\ f(x) = x(x – 1) (x – 2)

f ¢(x) = 3x² – 6x + 2 = 0

Ž x = 1 ± $\frac{1}{\sqrt{3}}$

\ The shown fig. has one positive & negative solution for

k = f$\left( 1 + \frac{1}{\sqrt{3}} \right)$

= $\left( 1 + \frac { 1 } { \sqrt { 3 } } \right)$ $\left( 1 + \frac{1}{\sqrt{3}} - 1 \right)$ $\left( 1 + \frac{1}{\sqrt{3}} - 2 \right)$

= $\frac{- 2}{3\sqrt{3}}$