Question

If the quadratic equation (1+a²)b²x²+ 2abcx + (c²-m²)=0 in x has equal roots, prove that c² = m² ( 1 + a²).​

Answer 1

Step-by-step explanation:

Given:-

• A quadratic equation (1+a²)b²x²+ 2abcx + (c²-m²) = 0.

• The roots of the equation are real and equal.

To Prove:-

• c² = m² ( 1 + a²)

Proof:-

For a quadratic equation ax² + bx + c = 0

If the roots of the equation are equal then:-

$\longrightarrow\boxed{ \bf b^2 - 4ac = 0}$

Comparing (1+a²)b²x²+ 2abcx + (c²-m²) with ax² + bx + c

• a = 1 + a²

• b = 2abc

• c = c² - m²

$\sf\longrightarrow b^2 - 4ac = 0$

$\sf\longrightarrow (2abc)^{2} - 4 \times \big[ (1 + {a)}^{2} {b}^{2} \big]\times ( {c}^{2} - {m}^{2}) = 0$

$\sf\longrightarrow4 {a}^{2} {b}^{2} {c}^{2} - 4 ( {b}^{2} + {a}^{2} {b}^{2})( {c}^{2} - {m}^{2}) = 0$

$\sf\longrightarrow\cancel4{a}^{2} {b}^{2} {c}^{2} = \cancel4 ( {b}^{2} + {a}^{2} {b}^{2})( {c}^{2} - {m}^{2})$

$\sf\longrightarrow{a}^{2} {b}^{2} {c}^{2} = ( {b}^{2} + {a}^{2} {b}^{2})( {c}^{2} - {m}^{2})$

$\sf\longrightarrow{a}^{2} {b}^{2} {c}^{2} = {b}^{2} {c}^{2} - {b}^{2} {m}^{2} + {a}^{2} {b}^{2}{c}^{2} - {a}^{2} {b}^{2} {m}^{2}$

$\sf\longrightarrow{a}^{2} {b}^{2} {c}^{2} - {a}^{2} {b}^{2} {c}^{2} = {b}^{2} {c}^{2} - {b}^{2} {m}^{2}- {a}^{2} {b}^{2} {m}^{2}$

$\sf\longrightarrow {b}^{2} {c}^{2} - {b}^{2} {m}^{2}- {a}^{2} {b}^{2} {m}^{2} = 0$

$\sf\longrightarrow {b}^{2} ({c}^{2} - {m}^{2}- {a}^{2} {m}^{2}) = 0$

$\sf\longrightarrow {c}^{2} - {m}^{2}- {a}^{2} {m}^{2}= 0$

$\sf\longrightarrow {c}^{2} - {m}^{2}(1+{a}^{2})= 0$

$\sf\dashrightarrow \underline{\boxed{\bf\therefore {c}^{2} - {m}^{2}(1+{a}^{2})= 0}}$

$\sf \underline{Hence, \: Proved}$

Points to be remembered while solving this kinda problems:-

For a quadratic equation ax² + bx + c = 0

If the roots are real and distinct then.

• b² - 4ac > 0

If the roots are real and equal

• b² - 4ac = 0

If the roots are imaginary.

• b² - 4ac < 0