Question

Find the roots of the quadratic equation (b-c) $x^2$+ (c-a) x +(a-b) =0.


can you please solve this without using the formula.
$take \: one \: root \: as \: \alpha \: and \: the \: other \: as \beta \: and \: then \: solve \: by \: taking \: sum \: and \: product \: of \: roots \:$
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Answer 1

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

➢ Given quadratic equation is

$\rm :\longmapsto\:(b - c) {x}^{2} + (c - a)x + (a - b) = 0$

By hit and trial,

If we put x = 1, then we get

$\rm :\longmapsto\:(b - c) {(1)}^{2} + (c - a)(1) + (a - b) = 0$

$\rm :\longmapsto\:(b - c)+ (c - a) + (a - b) = 0$

$\rm :\longmapsto\:b - c+ c - a+ a - b = 0$

$\rm :\longmapsto\:0 = 0$

$\bf\implies \:x = 1 \: is \: one \: root \: of \: given \: equation.$

$\rm :\longmapsto\:Let \: assume \: that \: other \: root \: be \: \alpha .$

We know, in a quadratic equation,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\:1 \times \alpha = \dfrac{a - b}{b - c}$

$\rm :\longmapsto \: \alpha = \dfrac{a - b}{b - c}$

Hence,

The roots of quadratic equation

$\bf :\longmapsto\:(b - c) {x}^{2} + (c - a)x + (a - b) = 0$

are

$\bf :\longmapsto \: 1, \: \dfrac{a - b}{b - c}$

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