Question

$if \: the \: roots \: of \: the \: equation \: {x}^{2} + 2cx + ab = 0 \: are \: real \: and \: unequal.then \: prove \: that \: the \: equation \: {x}^{2} - (a + b)x + {a}^{2} + {b}^{2} + 2 {c}^{2} = 0 \: has \: no \: real \: roots$

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Answer 1

$\huge{\mathcal{Hi\: there!}}$

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Given :

$\mathsf{ x {}^{2} + 2cx + ab = 0}$

has real and unequal roots, then

$\mathsf{b {}^{2} - 4ac > 0} \\ \\ \mathsf{(2c) {}^{2} - 4(ab) > 0} \\ \\ \mathsf{4c {}^{2} - 4ab > 0} \\ \\ \mathsf{c {}^{2} > ab \: ........(1) }$

Now,

$\mathsf{x {}^{2} - 2(a + b)x + a {}^{2} + b {}^{2} + 2c {}^{2} = 0 }$

Find the value of b² - 4ac :

⇒ {2 ( a + b )}² - 4 ( a² + b² + 2c² ) = 0

⇒ 4 ( a² + b² + 2ab ) - 4 ( a² + b² + 2c² ) = 0

⇒ 4a² + 4b² + 8 ab - 4a² - 4b² - 8c² = 0

⇒ 8 ( ab - c² )

From (1), we know that ;

$\mathsf{c {}^{2} > ab }$

So, roots of the second equation are less than 0.

$\implies \mathsf{b {}^{2} - 4ac < 0}$

∴ the equation

$\mathsf{x {}^{2} - 2(a + b)x + a {}^{2} + b {}^{2} + 2c {}^{2} = 0 }$

has no real roots.

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Thanks for the question !

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Answer 2

(i)

Given Equation is x^2 + cx + ab = 0

Here, a = 1, b = 2c, c = ab.

It has real and unequal roots.

⇒ D = b^2 - 4ac > 0

   = (2c)^2 - 4(1)(ab) > 0

   = 4c^2 - 4ab > 0

   = c^2 - ab > 0

(ii)

Given Equation is x^2 - 2(a + b)x + a^2 + b^2 + 2c^2 = 0

Here, a = 1, b = -2(a + b), c = a^2 + b^2 + c^2

It has no real roots.

⇒ D = b^2 - 4ac < 0

   = (-2[a + b]^2) - 4(1)(a^2 + b^2 + c^2) < 0

   = 4(a^2 + b^2 + 2ab) - 4(a^2 + b^2 + 2c^2) < 0

   = 8ab - 8c^2 < 0

   = 8(ab - c^2) < 0

   = -8(c^2 - ab) > 0

   From (i) c^2 - ab > 0.

Therefore, the given equation has no real roots.

Hope it helps!