Quadratic Equations

Samia CHALLAL

18 Quadratic Equations

The equation $a x^2 + bx + c = 0$ , with $a\not= 0$ , is called a quadratic equation.

The discriminant of this equation is given by $\Delta = b^2 - 4 a c$ .

The equation has:

– one solution if $\Delta =0$ , given by : $\displaystyle{ x= \frac{-b}{2a} }$

– two solutions if $\Delta >0$ , given by:

$\displaystyle{ x= \frac{-b + \sqrt{\Delta} }{2a}}\qquad \qquad\hbox{ and } \qquad\qquad \displaystyle{ x= \frac{-b - \sqrt{\Delta} }{2a}}$

– no solution if $\Delta < 0$ .

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How to complete a square?

$\displaystyle{ a x^2 + bx + c\quad = \quad a \Big( x^2 + \frac{b}{a} x \Big) + c \quad =\quad a \Big( x^2 +2 \frac{b}{2a} x +\big(\frac{b}{2a} \big)^2 - \big(\frac{b}{2a} \big)^2 \Big) +c }$

$\displaystyle{ \qquad \qquad\qquad = \quad a\Big( \big[ x + \frac{b}{2a} \big]^2 - (\frac{b}{2a} )^2 \Big) +c \quad = \quad a \big[ x + \frac{b}{2a} \big]^2 - a (\frac{b}{2a} )^2 + c }$

$\displaystyle{ \qquad \qquad \qquad = \quad a \big[ x + \frac{b}{2a} \big]^2 - \frac{b^2 - 4 ac }{4a} \quad = \quad a \Big( \big[ x + \frac{b}{2a} \big]^2 - \frac{b^2 - 4 ac }{4a^2} \Big) }$

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Show Examples

Exercise 1

Exercise 2

Show/Hide Solution.

The discriminant of the quadratic equation ${-6} x^2 + x + ({5})=0$ , is given by

$\Delta = (1)^2 - 4\Big( {-6}\Big) \Big(({5})\Big) = {121}$ .

The equation has two distinct solutions, given by :

$\displaystyle{ x= \frac{ -1 + \sqrt{ {121}} }{ {-12} } } =\frac{ 10 }{ -12 } =-\frac{ 5 }{6 } } \qquad$ $\quad$ and $\quad$ $\qquad\displaystyle{ x= \frac{-1 - \sqrt{{121}} }{{-12} } = 1.}$

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