Question

if two roots of a quadratic equation
$(1 + {m}^{2} ) {x}^{2} + 2mxc + ( {c }^{2} - {a}^{2} ) = 0$
are real and equal ,then let us prove that
${c }^{2} = {a}^{2} (1 + {m}^{2} )$
​

Answer 1

Step-by-step explanation:

Given:-

• A quadratic equation (1 + m²)x² + 2mxc + (c² - a²) = 0.

• The roots of the quadratic equation are real and equal.

To Prove:-

• c² = a²(1 + m²)

Concept used:-

For a quadratic equation ax² + bx + c = 0

If the roots are real and equation then,

$\sf \dashrightarrow \boxed{\sf b^2 - 4ac = 0}$

Solution:-

Comparing (1 + m²)x² + 2mxc + (c² - a²) = 0 with ax² + bx + c = 0

Here:-

• a = 1 + m²

• b = 2mc

• c = c² - a²

$\sf \implies {b}^{2} - 4ac = 0$

$\sf \implies {(2mc)}^{2} - \big[4( {c}^{2} - {a}^{2})(1 + {m}^{2}) \big] = 0$

$\sf \implies {(2mc)}^{2} = 4( {c}^{2} - {a}^{2})(1 + {m}^{2})$

$\sf \implies4 {m}^{2} {c}^{2} = 4( {c}^{2} - {a}^{2})(1 + {m}^{2})$

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

$\sf \implies \cancel{{m}^{2} {c}^{2} } = {c}^{2} + \cancel{{m}^{2} {c}^{2} } - {a}^{2} - {a}^{2} {m}^{2}$

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

$\sf \implies {c}^{2} = {a}^{2} + {a}^{2} {m}^{2}$

$\sf \implies {c}^{2} = {a}^{2}(1 + {m}^{2} )$

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

Hence Proved.

Answer 2

Given-

• $\bf(1 + {m}^{2} ) {x}^{2} + 2mxc + ( {c }^{2} - {a}^{2} ) = 0$
• In the above equation,roots are equal and real

To prove-

$\bf{c }^{2} = {a}^{2} (1 + {m}^{2} )$

Concept used-

• $\bf \: d = {b}^{2} - 4ac$
• and d=0

Proof-

We know,

$\bf \: d = {b}^{2} - 4ac$

Here,

$\bf\rightarrow \: a = 1 + {m}^{2} \: \: \: \: \: \\ \bf \rightarrow \: b = 2mc \: \: \: \: \: \: \: \: \: \: \\ \rightarrow \bf \: c = {c}^{2} - {a}^{2} \: \: \: \: \: \\ \because \bf roots \: are \: equal \: \\ \bf \therefore \: d = 0 \\ \\ \bf \: 0 = {(2mc)}^{2} - 4(1 + {m}^{2})( {c}^{2} - {a}^{2} ) \: \: \: \: \: \\ \\ \implies\bf \: 4( {c}^{2} - {a}^{2} + {m}^{2} {c}^{2} - {m}^{2} {a}^{2}) = 4 {m}^{2} {c}^{2} \\ \\ \implies \bf {c}^{2} - {a}^{2} + {m}^{2} {c}^{2} - {m}^{2} {a}^{2} = {m}^{2} {c}^{2} \: \: \: \: \: \: \: \: \: \\ \\ \bf \implies \: {c}^{2} - {a}^{2} - {m}^{2} {a}^{2} = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf \implies {c}^{2} = {a}^{2} + {m}^{2} {a}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\underline{ \boxed{ \huge{ { \bf \: c}^{2} = {a}^{2} (1 + {m}^{2} )}}} \\ \\ \bf \underline{hence \: proved}$