Question

Solve the following quadratic

${x}^{2} + ( \frac{c}{c + d} + \frac{c + d}{d} )x + 1 =$
​

Answer 1

Given Question :-

Solve the following quadratic equation :-

$\rm \: {x}^{2} + \bigg(\dfrac{c}{c + d} + \dfrac{c + d}{c} \bigg)x + 1 = 0$

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm \: {x}^{2} + \bigg(\dfrac{c}{c + d} + \dfrac{c + d}{c} \bigg)x + 1 = 0$

can be rewritten as

$\rm \: {x}^{2} + \bigg(\dfrac{c}{c + d} + \dfrac{c + d}{c} \bigg)x +\dfrac{c}{c + d} \times \dfrac{c + d}{c} = 0$

can be further rewritten as

$\rm \: {x}^{2} + \dfrac{c}{c + d} x + \dfrac{c + d}{c}x +\dfrac{c}{c + d} \times \dfrac{c + d}{c} = 0$

$\rm \: x\bigg(x + \dfrac{c}{c + d} \bigg) + \frac{c + d}{c}\bigg(x + \dfrac{c}{c + d} \bigg) = 0 \:$

$\rm \: \bigg(x + \dfrac{c}{c + d} \bigg) \bigg(x + \dfrac{c + d}{c} \bigg) = 0 \:$

So,

$\rm\implies \:x \: = \: - \: \dfrac{c}{c + d} \: \: or \: \: \: x \: = \: - \: \dfrac{c + d}{c}$

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Short Cut Trick :-

The quadratic equation of the form

$\rm \: {x}^{2} + \bigg(a + \frac{1}{a} \bigg)y + 1 = 0 \:$

always have the solution

$\: \boxed{\sf{ \: \: x \: = \: - \: a \: \: \: or \: \: \: x \: = \: - \: \frac{1}{a} \: \: }}$

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Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac