Question

6Q If the roots of the equation
(a²+b²) x² - 2(ab+cd)x+(c^2+d^2)=0 are
equal Prove that a/b=c/d​

Answer 1

Correct Question

If the roots of the equation  $( a^2 + c^2 ) x^2 - 2( ab + cd ) x + ( b^2 + d^2 ) = 0$ are equal.

 

Calculation Method

The discriminant of the quadratic equation $D/4=b'^2-ac$*⁽¹⁾ should equal zero.

$\implies D/4 = ( ab + cd )^2 - ( a^2 + c^2 ) ( b^2 + d^2 )$

Some of you might have seen this before if you had learned Cauchy inequality.*⁽²⁾

This actually evaluates to

$\implies D/4 = - ( ad - bc ) ^ 2$

$\therefore ( ad - bc ) ^ 2 = 0 \implies ad - bc = 0$

$\therefore ad = bc$

 

Note that

But $ad = bc$ doesn't always mean $\dfrac{a}{b} =\dfrac{c}{d}$, as the denominator cannot be 0.

$\therefore \dfrac{a}{b} = \dfrac{c}{d} \: ( b, d \neq 0 )$

 

More information

⁽¹⁾If you find discriminant on the quadratic equation $ax^2+2b'x+c=0$, the discriminant is actually $D = 4 ( b'^2 - ac )$. This is a special case where the x coefficient is even.

⁽²⁾Calculation includes $( a^2 + c^2 ) ( b^2 + d^2 ) - ( ab + cd )^2 = ( ad - bc )^2$.

Answer 2

Question :-

If the roots of the equation (a² + c²)x² - 2(ab + cd)x + (b² + d²) then prove that -

• $\sf \frac{a}{b} = \frac{c}{d}$

Given :-

• Roots of the quadratic equation are equal.

Solution :-

Comparing the given equation with ax² + bx + c = 0

$\sf \star \: a = ( {a}^{2} + {c}^{2} ) \\ \\ \sf \star \: b = 2(ab + cd) \\ \\ \sf \star \: c = ( {b}^{2} + {d}^{2} )$

We know that in a quadratic equation the discriminant :-

• $\sf \: \: d = {b}^{2} - 4ac$

But given that roots are equal so the discriminant will be 0. So,

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

Now,

$\sf { \big(2(ab + cd) \big)}^{2} - 4 \times ( {a}^{2} + {c}^{2} ) \times ( {b}^{2} + {d}^{2} ) = 0 \\ \\ \sf \implies \: 4 \{( {ab + cd)}^{2} - ( {a}^{2} + {c}^{2}) \times ( {b}^{2} + {d}^{2} )\} = 0 \: \\ \\ \sf \implies \: ( {ab + cd)}^{2} - ( {a}^{2} + {c}^{2}) \times ( {b}^{2} + {d}^{2} ) = 0 \\ \\ \sf \implies \: \cancel{ {a}^{2} {b}^{2}} + \cancel{ {c}^{2} {d}^{2}} + 2abcd - \cancel{{a}^{2} {b}^{2}} - {a}^{2} {d}^{2} - {b}^{2} {c}^{2} - \cancel{ {c}^{2} {d}^{2}} = 0 \\ \\ \sf \implies \: - \{ {a}^{2} {d}^{2} + {b}^{2} {c}^{2} - 2 \times ad \times bc \} = 0 \\ \\ \sf \implies - ( {ad - bc)}^{2} = 0 \\ \\ \sf \implies \: ( {ad - bc)}^{2} = 0 \\ \\ \sf \implies \: ad - bc = 0 \\ \\ \sf \implies \: ad = bc \\ \\ \implies \boxed{ \sf \: \frac{a}{b} = \frac{c}{d} }$

Hence proved !!

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Hope it will help uhh !!