Question

if (1 + a square) b Square X square + 2 ab x c square-m square =0 has equal roots show that c square =m square (1 + a square)​

Answer 1

Correct Question

If quadratic equation (1 + a²)b²x² + 2abcx + c² - m² = 0 has equal roots, show that c² = m²(1 + a²)

Solution

Since the quadratic equation has equal roots, the discriminant should equal to 0

Discriminant = B² - 4AC = 0

Comparing the given equation with standard equation Ax² + Bx + C, we get

A = (1 + a)²b²

B = 2abc

C = c² - m²

So,

(2abc)² - 4(1 + a²)b²(c² - m²) = 0

→ 4a²b²c² = 4(1 + a²)b²(c² - m²)

Cancel 4 and b² from both sides

→ a²c² = (1 + a²)(c² - m²)

→ a²c² = c² + a²c² - m² - a²m²

Cancel a²c² from both sides

→ c² - m²(1 + a²) = 0

→ c² = m²(1 + a²)

Answer 2

$\huge\underline\mathrm{Correct\:Question-}$

If (1+a²)b²x² + 2abcx + c² - m² = 0 has equal roots, then show that c² = m²(1+a²)

$\huge\underline\mathrm{Solution-}$

It is given that equation has equal roots.

$\therefore$ Discriminant = 0

: $\implies$ b² - 4ac = 0

___________

Comparing with ax² + bx + c = 0, we get,

• a = (1+a²)b²

• b = 2abc

• c = c² - m²

Putting the values,

: $\implies$ (2abc)² - 4(1+a²)b²(c²-m²) = 0

: $\implies$ $\sf{\cancel{4b^2}}$ a²c² = $\sf{\cancel{4b^2}}$ (1+a²)(c²-m²)

: $\implies$ $\sf{\cancel{a^2\:c^2}}$ = c² + $\sf{\cancel{a^2\:c^2}}$ - m² - a²m²

: $\implies$ m²(1-a²) = c²

$\large{\boxed{\red{\rm{c^2\:=\:m^2\:(1-a^2)}}}}$

Hence proved!