The general solution to a quadratic equation

Quadratic Equations: Quadratic Equations with one Unknown

The general solution to a quadratic equation

Assume that #a#, #b#, and #c# are real numbers with #a\neq 0#.

The discriminant of the quadratic equation #ax^2+bx+c = 0# with unknown #x# is the number #b^2-4\cdot a \cdot c#.

The sign of the discriminant is defining for the number of solution to the equation.

If we divide the equation by #a#, we get to the equivalent equation #x^2+\frac{b}{a}x+\frac{c}{a} = 0#, then the discriminant is equal to #\left(\frac{b}{a}\right)^2-4\cdot1\cdot\frac{c}{a}=\frac{1}{a^2}\left(b^2-4\cdot a\cdot c\right)#. Hence the sign of the discriminant does not change.

As with the linear case we will also consider equations as quadratic equations that are quadratic after we have factored out the brackets and moved all terms to the left hand side.

The formula below, which gives us the solutions directly, is called the abc-formula. The discriminant plays a crucial role in deciding how many solutions there are.

The quadratic equation #ax^2+bx+c = 0# with unknown #x# and discriminant #d=b^2-4ac# has:

no solutions if #d\lt 0#
exactly one solution if #d=0#, being #x=-\dfrac{b}{2a}#
two solutions if #d\gt 0#, namely #x=\dfrac{-b - \sqrt{d}}{2a}# and #x=\dfrac{-b+ \sqrt{d}}{2a}#.

By completing the square it is possible to rewrite the equation as #\displaystyle a\left(x+\frac{b}{2a}\right)^2 -\frac{b^2-4ac}{4a}=0#. If we move the constant term to the right hand side and divide by #a#, we find \[\left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2-4ac}{(2a)^2}\tiny.\]If #b^2-4ac\lt 0#, the right hand side is negative, but the left hand side is not, so there are no solutions.

If #b^2-4ac= 0#, the right hand side, and therefore also the left hand side, are equal to #0#, after which follows that #x=-\dfrac{b}{2a}# is the only solution.

If #b^2-4ac\gt 0#, we can take the root; these two roots are, apart from one sign, equal to each other: #x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{b^2-4ac}}{2a}#. Hence, there are two solutions, #x=\dfrac{-b - \sqrt{b^2-4ac}}{2a}# and #x=\dfrac{-b+ \sqrt{b^2-4ac}}{2a}#.

The two solution in the last case are often taken together by using the notation #\pm#; therefore #x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}#.

If #a=0#, it follows that #bx+c=0#, hence a linear equation that can also be solved. Therefore each equation of the form #ax^2+bx+c=0# can be solved.

Solve #x# in the quadratic equation #2\cdot x^2+6\cdot x+4= 0#.

Give your answer in the form #x=a\vee x=b# if there are two solutions #x=a# if there is one solution and #none# if there are no solutions.

# x=-2\vee x=-1#

We solve the equation by completing the square:\[\begin{array}{rclcl}2\cdot x^2+6\cdot x+4 &=&0&\phantom{x}&\color{blue}{\text{the original equation}}\\ 2(x+{{3}\over{2}})^2-{{1}\over{2}} &=&0&\phantom{x}&\color{blue}{\text{the square completed}}\\ (x+{{3}\over{2}})^2 &=&{{1}\over{4}}&\phantom{x}&\color{blue}{\text{constants to the right}}\\ (x+{{3}\over{2}}) &=&\pm\sqrt{{{1}\over{4}}}&\phantom{x}&\color{blue}{\text{root extracted}}\\ x&=&-{{3}\over{2}}\pm\sqrt{{{1}\over{4}}}&\phantom{x}&\color{blue}{\text{constant to the right}}\end{array}\] Hence, the answer is: #x=-2\vee x=-1#.

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