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Modulus of a Complex Number

If the equati...

Question

If the equation $(m^{2} + n^{2}) x^{2} - 2 (m p + n q) x + p^{2} + q^{2} = 0$ has equal roots , then

m p = n q

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m q = n p

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m n = p q

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m q = \sqrt{n p}

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Solution

The correct option is B $m q = n p$
For real and equal root, discriminant, $D = 0$
$b^{2} = 4 a c$
$\Rightarrow 4 (m p + n q)^{2} = 4 (m^{2} + n^{2}) (p^{2} + q^{2})$
$\Rightarrow m^{2} q^{2} + n^{2} p^{2} - 2 m n p q = 0$
$\Rightarrow (m q - n p)^{2} = 0 \Rightarrow m q - n p = 0$
$∴ m q = n p$

Suggest Corrections

Similar questions

On dividing $2 m p^{2} q - m q p + 4 n p q^{2}$ by $p q$ , what do we get?

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

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{t a n}^{- 1} \frac{2 m n}{m^{2} - n^{2}} + {t a n}^{- 1} \frac{2 p q}{p^{2} - q^{2}} = {t a n}^{- 1} \frac{2 M N}{M^{2} - N^{2}}

M = m p - n q, N = n p + m q

\frac{2}{3} {t a n}^{- 1} \frac{3 m n^{2} - m^{3}}{n^{3} - 3 m^{2} n} + \frac{2}{3} {t a n}^{- 1} \frac{3 p q^{2} - p^{3}}{q^{3} - 3 p^{2} q} = {t a n}^{- 1} \frac{2 M N}{M^{2} - N^{2}}

M = - m p + n q, N = n p + m q

+ i v e

If Δ A B C \sim Δ M N P, A D ⊥ B C

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\frac{A D}{M Q} = \frac{3}{6}

∠ N

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