Quadratic Formula Example with rational roots:
*Solve the equation x2 + 3x – 4 = 0
using quadratic formula?
If we find its factors, we have (x+4)(x-1)=0, which yields -4 and 1 as the roots of equation. Since we are solving quadratic formula examples, let's do it with quadratic formula. Comparing it with ax2 + bx +c = 0
, we have a=1, b=3, c=-2. Applying the formula,
Solving it for both positive and negative values, we have x=4 and 1 which we obtained by factoring the equation. If there are rational roots, it's easy to acquire solution. But think the other case. Let's have an quadratic formula example with irrational roots.
Quadratic Formula example without rational roots:
*Next, let's solve the equation x2 + 3x – 2 = 0 using quadratic formula?
Here, if you try to factoring it, it's a cumbersome process. You need to go for quadratic formula method which is given below. Here we have a=1, b=3 and c=-2. Substituting all this values in the formula for quadratic equation,
which are not rational roots. We can approximate the results using calculator which would come around 0.56 and -3.56.
Example for quadratic equation having complex roots:
If the discriminant value i.e. b2-4ac < 0, than it's square-root yields complex numbers and hence the roots of the quadratic equation are also complex. Let's see solution to one such equation.
*Solve the equation x2+4x+5=0 using quadratic equation?
Here we have a=1, b=4, c=5. Clearly b2-4ac = -4<0. We know that square-root of -1 is 'i'. Applying the formula, we have
which results in 2 complex roots, -2+2i and -2-2i. Thus the formula provides a effective solution to all kinds of quadratic equations.
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