Nature of Roots Solution5

In this page nature of roots solution5 we are going to see solutions of some practice questions of the worksheet nature of roots.

(6) Show that the roots of the equation (x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a) = 0 are always real and they cannot be equal unless a = b = c.

To prove that the roots of the given equal is always real we have to find the quadratic equation.

(x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a) = 0

for that we have to multiply the terms.

x² - b x - a x + a b + x² - c x - b x + b c + x² - a x - c x + a c = 0

x² + x² + x² - b x - a x - c x - b x - a x - c x + a b + b c + a c = 0

3 x² - 2 b x - 2 a x - 2 c x + a b + b c + c a = 0

3 x² + (- 2 a - 2 b - 2 c) x + (a b + b c + c a) = 0

a = 3          b  = (- 2 a - 2 b - 2 c)     c = (a b + b c + c a)

b² - 4 a c = (- 2 a - 2 b - 2 c)² - 4 (3) (a b + b c + c a)

= -2(a + b + c)² - 12 (a b + b c + c a)

= 4 (a²+b²+c²+2ab+2bc+2ca)-12 a b -12 b c - 12 c a

= 4 a²+ 4 b²+ 4 c²+ 8 ab+ 8 bc+ 8 ca-12 a b -12 b c - 12 c a

= 4 a²+ 4 b²+ 4 c² - 4 a b - 4 b c - 4 c a

= 4 (a²+ b²+ c² - a b - b c - c a)

now let us check what will happen if a = b = c

from this we come to know that we can apply "a" instead of "b" and "c"

= 4 (a²+ a²+ a² - a (a) - (a) (a) - (a) (a))

= 4 (a²+ a²+ a² - a² - a² - a²)

= 0

if we take a = b = c  then we get ∆ = 0. From this we can decide that the roots are always real and they cannot be equal unless a = b = c.

(7) If the equation (1 + m²) x² + 2 m c x + c² - a² = 0 has equal roots, then prove that c² = a²(1 + m²)

Since the roots of the equation are real then ∆ = 0

∆ = b² - 4 a c

a = (1 + m²)        b = 2 m c        c = c² - a²

(2 m c)² - 4 (1 + m²) (c² - a²) = 0

4 m² c² - 4(c² - a² + m² c² - m² a²) = 0

4 m² c² - 4c² + 4 a² - 4 m² c² + 4 m² a² = 0

4 a² - 4c² + 4 m² a² = 0

4 a² + 4 m² a² =  4c²

dividing by 4 on both sides

a² + m² a² =  c²

a²(1 + m²) = c²

c² =  a²(1 + m²)

Hence proved nature of roots solution5 nature of roots solution5