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Question

Number of roots of a quadratic equation  are:


A
1
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B
2
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C
3
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D
None
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Solution

The correct option is A $$2$$

Let us, consider the quadratic equation of the general form  

$$ax^{2}+bx+c=0$$  where, $$a≠0$$

Now divide each term by $$a$$ (Since  $$a≠0$$), we get 

$$\Rightarrow$$ $$x^{2}$$ $$+$$ $$\dfrac{b}{a}x$$$$+$$$$\dfrac{c}{a}=0$$

$$\Rightarrow$$ $$x^{2}$$$$+$$$$\dfrac{2b}{2a}x$$ $$+$$ $$\Bigg(\dfrac{b}{2a}\Bigg)^{2}$$$$-$$$$\Bigg(\dfrac{b}{2a}\Bigg)^{2}$$$$+$$$$\dfrac{c}{a}$$  = $$0$$ 

After this we get

$$\Rightarrow$$ $$\Bigg(x + \dfrac{b}{2a}\Bigg)^{2}$$ – $$\Bigg(\dfrac{\sqrt{b^{2} - 4ac}}{2a}\Bigg)^{2}=0$$

$$\Rightarrow$$ $$\Bigg[x + \Bigg(\dfrac{b}{2a}\Bigg) + \Bigg(\dfrac{\sqrt{b^{2} - 4ac}}{2a}\Bigg)\Bigg]\Bigg[x +
\Bigg(\dfrac{b}{2a}\Bigg) - \Bigg(\dfrac{\sqrt{b^{2} - 4ac}}{2a}\Bigg)\Bigg]=0$$

$$\Rightarrow$$ $$\Bigg[x - \Bigg(\Bigg(\dfrac{-b - \sqrt{b^{2} - 4ac}}{2a}\Bigg)\Bigg)\Bigg]\Bigg[x - \Bigg(\Bigg(\dfrac{-b +
\sqrt{b^{2} - 4ac}}{2a}\Bigg)\Bigg)\Bigg] = 0$$

$$\Rightarrow$$ $$(x - α)(x - β) = 0, $$

where $$α = \Bigg(\dfrac{- b -\sqrt{b^{2} - 4ac}}{2a}\Bigg)$$ 

and $$β = \Bigg(\dfrac{- b + \sqrt{b^{2} - 4ac}}{2a}\Bigg)$$

Now we can clearly see that the equation $$ax^{2} + bx + c = 0$$ reduces to
$$(x - α)(x - β) = 0$$ and the equation $$ax^{2} + bx + c = 0$$ is only satisfied
by the values $$x = α$$ and $$x = β$$.

Except $$α$$ and $$β$$ no other values of x satisfies the equation

$$ax^{2} + bx +c = 0$$

Hence, we can say that the equation $$ax^{2} + bx + c = 0$$ has $$only$$ $$two$$ roots.

Therefore, number of roots of Quadratic Equation are $$2$$

Hence Answer is $$(B)$$ 


Mathematics

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