Question

Ello♥️♥️♥️
if the roots of the equation
$( {a}^{2} + {b}^{2} ) {x}^{2} - 2(ac + bd)x + ( {c}^{2} + {d}^{2} ) = 0$
Are equal then prove that

$\frac{a}{b} = \frac{c}{d}$

​

Answer 1

Given Quadratic equation,

$( {a}^{2} + {b}^{2} ) {x}^{2} - 2(ac + bd)x + ( {c}^{2} + {d}^{2} ) = 0$

We know that for a quadratic equation in x ( ax² + bx + c = 0) , The roots are given by,

$x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

So clearly We get equal roots when b² = 4ac.

For the given question,

Coefficient of x² = a² +b²

Coefficient of x =-2 ( ac + bd)

Constant = c² + d²

So We have,

(-2)²(ac+bd)² = 4(a² + b²) (c²+d²)

a²c² + b²d² + 2abcd = ( a²c² + b²c²+a²d² + b²d²)

2abcd = a²d² + b²c²

a²d² - abcd - abcd + b²c² = 0

ad ( ad - bc) - bc ( ad - bc) =0

(ad-bc)² = 0

Finally, ad - bc = 0

$ad = bc \\ \\ \frac{a}{b} = \frac{c}{d}$

Hence proved.

Answer 2

Answer:

$\sf \dfrac{a}{b}=\dfrac{c}{d}$  

Hence proved.

Step-by-step explanation:

Given :

(a² + b²)x² - 2(ab + cd)x + (c² + d²) = 0.

To Prove :

$\sf \dfrac{a}{b}=\dfrac{c}{d}$  

Proof :

As Given,

(a² + b²)x² - 2(ab + cd)x + (c² + d²) = 0.

Here,

Roots are equal.

So,

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

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

Where,

$\sf\\ \\\implies a = a^2+b^2\\ \\\implies b= -2(ac+bd)\\ \\\implies c=c^2+d^2$

We know that :

$\bigstar\;\boxed{\sf -2(ac+bd)^2=4(a^2+b^2)(c^2+d^2)}}$

So,

$\sf \\ \\ \implies 4a^2c^2+8acbd+4b^2d^2\\ \\\implies 4a^2c^2+4a^2d^2+4b^2c^2+4b^2d^2\\ \\ \implies 4a^2c^2\; and\;4b^2b^2 d\quad[Here\;Get\; Cancelled]$

Now,

$\sf\\ \implies4a^2d^2+4b^2c^2-8acbd=0$

Here, Taking the number "4" as common.

$\sf\\ \implies4(a^2d^2-8acbd+b^2c^2)=0$

According to the Identity,

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

_________{ Taking square root at L.H.S and R.H.S! }

$\sf\\ \implies d-bc=0\\ \\ \implies ad=bc\\ \\\implies \dfrac{a}{b}=\dfrac{c}{d}$

Hence, Proved.