Question

write a condition fot nature of roots of a quadratic equation ax^ 2+ bx + c = 0​

Answer 1

If b^2 -4ac =0

then the roots are equal and real.

If b^2 -4ac <0

then the roots are not real .

If b^2 -4ac >0

then the roots are real and unequal

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Answer 2

AnsWer :

# How to Find Nature of Roots.

$\begin{tabular}{ | c | c | } \cline{1-2}When & Then \\ \cline{1-2} D more than 0 & The roots real \\ \cline{1-2} D = 0 & both roots are equal and Real roots. \\ \cline{1-2} D less than 0 & The roots are unreal and imaginary \\ \cline{1-2} \end{tabular}$

Point Note,

• Discriminant ( D ) It show nature of roots.
• D = b² - 4ac.

Let Find both Roots by Quadratic Formula.

We have,

General Equation,

$\tt \dagger \: \: \: \: \: a{x}^{2} + bx + c = 0.$

And, dividing coefficient of x² both sides.

$\tt\longmapsto \dfrac{a {x}^{2} }{ a } + \dfrac{bx}{ a} + \dfrac{c}{a} = 0.$

$\tt\longmapsto {x}^{2} + \dfrac{bx}{ a} + \dfrac{c}{a} = 0.$

Now, By completing the Square method

$\tt\longmapsto {x}^{2} + \dfrac{bx}{ a} + {( \dfrac{b}{2a}) }^{2} - {( \dfrac{b}{2a}) }^{2} + \dfrac{c}{a} = 0.$

$\tt \longmapsto {(x + \dfrac{b}{2a} )}^{2} - {(\dfrac{b}{2a})}^{2} + \dfrac{c}{a} = 0.$

$\tt\longmapsto {(x + \dfrac{b}{2a} )}^{2} = \dfrac{ {b}^{2} }{4 {a}^{2} } - \dfrac{c}{a}$

$\tt\longmapsto {(x + \dfrac{b}{2a} )} = \pm \sqrt{ \frac{ {b}^{2} - 4ac}{4 {a}^{2} } }$

$\tt\longmapsto {(x + \dfrac{b}{2a} )} = \pm \frac{\sqrt{ {b}^{2} - 4ac} }{2a}$

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

$\tt\longmapsto x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$\rule{90}1$

$\tt\longmapsto \alpha = \dfrac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a}$

$\tt\longmapsto \beta = \dfrac{ -b-\sqrt{ {b}^{2} - 4ac } }{2a}$