Question

Derive the Quadratic Formula- Class 10 Ch.4 Maths
Plz.. Don't spam I'm using all my points here :)

Answer 1

1. Complete the Square

ax2 + bx + c has "x" in it twice, which is hard to solve.

But there is a way to rearrange it so that "x" only appears once. It is called Completing the Square (please read that first!).

Our aim is to get something like x2 + 2dx + d2, which can then be simplified to (x+d)2

So, let's go:

Start with ax^2 + bx + c=0

Divide the equation by a x^2 + bx/a + c/a = 0

Put c/a on other side x^2 + bx/a = -c/a

Add (b/2a)2 to both sides x^2 + bx/a + (b/2a)^2 = -c/a + (b/2a)^2

The left hand side is now in the x2 + 2dx + d2 format, where "d" is "b/2a"

So we can re-write it this way:

"Complete the Square" (x+b/2a)^2 = -c/a + (b/2a)^2

Now x only appears once and we are making progress.

2. Now Solve For "x"

Now we just need to rearrange the equation to leave "x" on the left

Start with (x+b/2a)^2 = -c/a + (b/2a)^2

Square root (x+b/2a) = (+-) √(-c/a+(b/2a)^2)

Move b/2a to right x = -b/2a (+-) √(-c/a+(b/2a)^2)

That is actually solved! But let's simplify it a bit:

Multiply right by 2a/2a x = [ -b (+-) √(-(2a)^2 c/a + (2a)^2(b/2a)^2) ] / 2a

Simplify: x = [ -b (+-) √(-4ac + b^2) ] / 2a

Which is the Quadratic formula we all know and love:

Quadratic Formula: x = [ -b (+-) √(b^2 - 4ac) ] / 2a

here is your answer

Answer 2

We can derive the quadratic formula by mainly two methods, by multiplying or by dividing a standard quadratic equation by the coefficient of x².

A standard quadratic equation may be written in the form of

ax² + bx + c = 0.

Let's start the derivation now.

By Dividing the standard quadratic equation by the coefficient of x², a in the case above.

$ax^2 + bx + c = 0\\\\\text{Dividing both sides by a, we have}\\\\\dfrac{1}{a}(ax^2 + bx + c) = \dfrac{1}{a}(0)\\\\\\\implies x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0\\\\\\\implies x^2 + 2(x) \left( \dfrac{b}{2a} \right) + \dfrac{c}{a} = 0\\\\\\\implies x^2 + 2(x) \left( \dfrac{b}{2a} \right) + \left( \dfrac{b}{2a} \right)^2 - \left( \dfrac{b}{2a} \right)^2 + \dfrac{c}{a} = 0$

$\text{Using (a+b)^2 = a^2 + b^2 + 2ab , considering the first terms in the above equation, we have}$

$\left( x + \dfrac{b}{2a} \right)^2 = -\dfrac{c}{a} + \left( \dfrac{b}{2a} \right)^2\\\\\\\implies \left( x + \dfrac{b}{2a} \right)^2 = -\dfrac{c}{a} + \dfrac{b^2}{4a^2}\\\\\\\implies \left( x + \dfrac{b}{2a} \right)^2 = \dfrac{b^2 -4ac}{4a^2}\\\\\\\implies x + \dfrac{b}{2a} = \pm \sqrt{\dfrac{b^2 -4ac}{4a^2}}\\\\\\\implies x + \dfrac{b}{2a} = \pm \sqrt{\dfrac{b^2 -4ac}{(2a)^2}}\\\\\\\implies x + \dfrac{b}{2a} = \dfrac{\pm \sqrt{b^2 - 4ac}}{2a}$

$\implies x = - \dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 - 4ac}}{2a}\\\\\\\implies \boxed{x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

By Multiplying the standard quadratic equation by the coefficient of x², a in the case above.

$ax^2 + bx + c = 0\\\\\text{Multiplying both sides by a, we have}\\\\a(ax^2 + bx + c) = a(0)\\\\\implies a^2x^2 + abx + ac = 0\\\\\implies (ax)^2 + 2(ax) \left( \dfrac{b}{2} \right) + ac = 0\\\\\\\implies (ax)^2 + 2(ax) \left( \dfrac{b}{2} \right) +\left( \dfrac{b}{2} \right)^2 - \left( \dfrac{b}{2} \right)^2 + ac = 0\\\\\\\text{Using (a+b)^2 = a^2 + b^2 + 2ab ,}\\\\\text{Considering the first three terms in the above equation, we have}$

$\left( ax + \dfrac{b}{2} \right) ^2 = \dfrac{b^2}{2^2} - ac\\\\\\\implies \left( ax + \dfrac{b}{2} \right) ^2 = \dfrac{b^2}{4} - ac\\\\\\\implies \left( ax + \dfrac{b}{2} \right) ^2 = \dfrac{b^2 - 4ac}{4}\\\\\\\implies ax + \dfrac{b}{2} = \pm \sqrt{\dfrac{b^2 - 4ac}{4}}\\\\\\\implies ax + \dfrac{b}{2} = \dfrac{\pm \sqrt{b^2 - 4ac}}{2}\\\\\\\implies ax = - \dfrac{b}{2} \pm \dfrac{\sqrt{b^2 - 4ac}}{2}\\\\\\\implies ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2}$

$\implies \boxed{x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

These were the two ways using which you can derive Quadratic Formula.