Question

Show that $x = - \frac{bc}{ad}$ is a solution of the quadratic equation

$ad {}^{2} ( \frac{ax}{b} + \frac{2c}{d} )x + bc {}^{2} = 0.$

Answer 1

Hey there !!

The given quadratic equation :-


$ad {}^{2} ( \frac{ax}{b} + \frac{2c}{d} )x + bc {}^{2} = 0.$


$x = \frac{-bc}{ad}$  can be a Solution of the Given Quadratic Equation, If and only If, it satisfies the Equation.


Let us take the Given Quadratic Polynomial as p(x)

⇒ $p(x) = ad^2x(\frac{ax}{b} + \frac{2c}{d}) + B{c}^{2}$

We need to Prove :  $p(\frac{-bc}{ad}) = 0$

⇒ $p(\frac{-bc}{ad}) = ad^2( \frac{-bc}{ad})( \frac{a}{b}( \frac{-bc}{ad}) + \frac{2c}{d}) + bc^2$

⇒ $p(\frac{-bc}{ad}) = -dbc(\frac{-c}{d} + \frac{2c}{d}) + bc^2$

⇒ $p(\frac{-bc}{ad}) = -dbc(\frac{c}{d}) + bc^2$

⇒ $p(\frac{-bc}{ad}) = -bc^2 + bc^2$

⇒ $p(\frac{-bc}{ad}) = 0$

⇒ Therefore, $p(\frac{-bc}{ad}) = 0$

We can say that ( $x = \frac{ - bc}{ad}$ ) is a Solution of the Given Quadratic Equation.

✔✔ Hence, it is proved ✅✅.

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THANKS

#BeBrainly.

Answer 2

Given equation is ad^2(ax/b + 2c/d)x + bc^2 = 0

⇒ ad^2x(ax/b + 2c/d) + bc^2 = 0

⇒ (a^2d^2x^2/b) + (2acd^2x/d) + bc^2 = 0

⇒ (a^2d^2x^2/b) + (2acdx) + bc^2 = 0

⇒ a^2d^2x^2 + 2acd^2x + b^2c^2 = 0

It is in the form of a^2 + 2ab + b^2 = (a + b)^2

⇒ (adx + bc)^2 = 0

⇒ adx + bc = 0

⇒ adx = -bc

⇒ x = -bc/ad.

Hope it helps!