This page deals with solving quadratic equations using quadratic formula. Quadratic equations are simple to solve when the factors are obvious or even when there is just the squared part and the number part by using the square root method.
But when nothing looks obvious one can always depend on the quadratic formula. It works with all types of quadratic equations whatever be the coefficients a, b and c.
According to this method the roots of the equation:
aX2 + bX + c = 0
is given by
$$X=\frac{-b\pm\sqrt{b^2-4ac }}{2a}$$
The formula can be easily derived by the method of completing the square root which we did in the earlier section.
Care has to be taken while entering the values of the coefficient a, b and c. We will solve a few problems to understand the application better.
Quadratic Formula Problems
Solve the equation x2 + 5x – 4 = 0
Here there is no number in front of x2 , but then we can look at this as 1x2.
Hence here a = 1, b = 5, c = -4
Plugging these values in the quadratic formula problems we get:
$$ X=\frac{-5\pm\sqrt{5^2-4(1)(-4) }}{2(1)}$$
Simplifying the square root we get:
$$ X=\frac{-5\pm\sqrt{25+16 }}{2}$$
$$ X=\frac{-5\pm\sqrt{41 }}{2}$$
Roots of the equation are: $$-\frac{5}{2}\pm\frac{\sqrt{41}}{2}$$
The above equation cannot be factored hence the only alternative method here was completing the square root.
Online Quadratic Formula Solver
Solve the equation x(x-3) = 5
This equation cannot be factored nor can it be solved by quadratic formula directly.
First let’s bring it to the standard form by simplifying.
X2 – 3X = 5
X2 – 3X – 5 = 0
Now it is in the standard form.
Here a = 1, b = -3, c = -5
Use the quadratic formula: $$X=\frac{-b\pm\sqrt{b^2-4ac }}{2a}$$
$$X=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-5) }}{2(1)}$$
$$X=\frac{3\pm\sqrt{29}}{2}$$
Hence the solution is: $$X = \frac{3}{2}\pm\frac{\sqrt{29}}{2}$$
Quadratic Formula Solution
The expression inside the square root, b2 – 4ac, is called the "discriminant” because, by using its value, one can discriminate i.e. determine the various types of quadratic formula solutions.
Case I :
If b2 – 4ac = 0 then there will be just one solution. This means the graph of the quadratic will touch the x-axis only at one point.
Case II :
If b2 – 4ac > 0 then there will be two distinct solutions. The graph of the quadratic will cut the x-axis at 2 points.
Case III :
If b2 – 4ac < 0 then there will be no real roots. In this case the graph of the quadratic will not touch the x-axis at all.
