The formular method of the quadratic equation is often a popular request by student. So we have taken our time to break down the steps to how this formular was arrived at.
How the formular method was derived.
The formular method of the quadraric equation can be regarded as a derivative
Of completing the square method of solving the prototype of a quadratic equation.
I.e ax^2 + bx + c = 0
By solving the prototype quadratic equation above ,using completing the square method. We will arrive at the quadratic formular below
(-b +-_ /b^2-4ac)
2a
Now let's examine how the formular came about.
Take your equation (the quadratic equation prototype)
ax^2 + bx + c = 0
move the constant c in the quadratic equation to the R.H.S of the equation (R.H.S - right hand side)
we have :
ax^2 + bx = -c
now divide through by a
ax^2/a + bx/a = -c/a
we will arrive at :
x^2 + bx/a = -c/a
applying Completing the square method
x^2 + bx/a + (b/2a)^2 = - c/a + (b/2a)^2 ---------(1)
from the above equation,it could be pictured that half the co-efficient of x (of bx/a) which is b/2a was squared and addeto to both sides of the equation.(That is what we call completing the sqare method of the quadratic equation)
by solving equation (1):
we will have,
(x+b/2a)^2 = b^2/4a^2 - c/a
{note that (x+b/2a)^2 when expanded is the same as x^2 + bx/a +(b/2a)^2 }
Acting on the R.H.S of the equation,
(x+b/2a)^2 = b^2 - 4ac
4a^2
taking the square root of both sides we have,
x + b/2a = +-/b^2 - 4ac
2a
now solving for x we have,
X= -b/2a +- /b^2 - 4ac
2a
which could further be simplified as
X= -b+-/b^2-4ac
2a
i.e. X= -b+/b^2-4ac or X= -b-/b^2-4ac
2a 2a
That's just how the formula method of quadrtic equation was derived.
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