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A Quadratic Equation is any equation having the form
$$ax^2 + bx + c = 0$$ where
• • $x$ represents a variable or an unknown

• • a,b and c are constants

• • a ≠ 0

Here are some examples:
• $x^2 + 2x + 8 = 0$
• $4x^2 - 6x + 9 = 0$
• $x^2 - 25x = 0$
• $9x^2 + 81 = 0$

$9x + 5 = 0$ is not a quadratic equation as $x^2$ term is missing. It is a linear equation.

Few more points to remember-
• The quadratic equation involves only one unknown ($x$ in this case).
• The quadratic equation only contains powers of $x$ that are non-negative.
• The solution to the quadratic equation are those values of $x$ which make its value 0. These are called roots of the equation.

A quadratic equation with real or complex coefficients has two solutions, called roots.
These two solutions may or may not be distinct, and they may or may not be real.

One way to find the roots of a quadratic equation is by a method called 'Completing the square'.
Starting with a quadratic equation in a standard form $ax^2 + bx + c = 0$

• Divide each side by $a$
⇒ $x^2 + \frac{bx}{a} + \frac{c}{a} = 0$

• Subtract constant $\frac{c}{a}$ from both sides
⇒ $x^2 + \frac{bx}{a} = \frac{-c}{a}$

• Add the square of one-half of $\frac{b}{a}$, the coefficient of $x$, to both sides.
⇒ $x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} = \frac{-c}{a} + \frac{b^2}{4a}$

• Now we can re-write left side as a square and simplify right hand side a bit.
⇒ $(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a}$

• Take square root on both sides.
⇒ $x + \frac{b}{2a} = ± \frac{\sqrt{b^2 - 4ac}}{2a}$
⇒ $x = ± \frac{\sqrt{b^2 - 4ac}}{2a} - \frac{b}{2a}$

$$x = \frac{\text{-b ±\sqrt{b^2 - 4ac}}}{2a}$$ The solution(s) to a quadratic equation can be found by using this formula.

Notice the ± symbol. This means that both $\frac{\text{-b +$\sqrt{b^2 - 4ac}$}}{2a}$ and $\frac{\text{-b -$\sqrt{b^2 - 4ac}$}}{2a}$ are roots of the quadratic equation.

Let us now solve the following quadratic equation using this method.

$4x^2 - 12x + 3 = 0$

• Divide each side by 4
⇒ $x^2 - \frac{12x}{4} + \frac{3}{4} = 0$

• Subtract constant $\frac{3}{4}$ from both sides
⇒ $x^2 - \frac{12x}{4} = \frac{-3}{4}$
⇒ $x^2 - 3x = \frac{-3}{4}$

• Add the square of one-half of $3x$, the coefficient of $x$, to both sides.
⇒ $x^2 -3x + \frac{9}{4} = \frac{-3}{4} + \frac{9}{4}$

• Now we can re-write left side as a square and simplify right hand side a bit.
⇒ $(x - \frac{3}{2})^2 = \frac{6}{4}$

• Take square root on both sides.
⇒ $x - \frac{3}{2} = ± \frac{\sqrt{6}}{2}$
⇒ $x = ± \frac{\sqrt{6}}{2} + \frac{3}{2}$
⇒ $x = \frac{3 ± \sqrt{6}}{2}$
⇒ $x = \text{2.72474487 or 0.27525512}$

Geometric Interpretation

The graph of a quadratic equation has a shape called parabola.
The values of a, b and c determine the location and size of the parabola.

See how the graphs of quadratic equation and linear equation differ from each other.
 Graphical representation of Quadratic Equation Graphical representation of Linear Equation

The simplest quadratic equation is $f(x) = x^2$. This equation represents a parabola.
In this graphical representation, $f(x) = x^2$. In this case, a = 1.

As we change the value of a, see how the shape of parabola changes.
$g(x) = \frac{1}{2}x$ ⇒ For values of $a$ less than 1, the parabola tends to be flatter and it expands outwards.
$h(x) = 2x$ ⇒ For values of $a$ greater than 1,the parabola curves inwards.
$j(x) = - x^2$. ⇒ For values of $a$ less than 0, the parabola flips upside down.

Quadratic equations are very useful in everyday life, as when finding out the speed and height of an object, determining product's profit and calculating areas.
Let us look at one such example to find the speed of a boat which moves up and down the river.

Assume a boat is going up a river. The river moves at 2 km per hour. If the boat goes upstream against the current for 20 km, the trip takes it 4 hours to go there and return.
Let us try to find the speed of the boat in water using quadratic equation.

As we know time = $\frac{distance}{speed}$
Let $x$ be the boat's speed in the water.

While traveling upstream, the boat's speed is v = x - 2 due to the resistance from the river current flowing at 2 km/hour
While going downstream, the boat's speed is v = x + 2 as river current helps it move faster.

The total time taken is equal to 4 hours, which is equal to the time going upstream plus the time going downstream. The total distance travelled is 20 + 20 = 40 km.

So we know that 4 hours = $\frac{20}{x-2} + \frac{20}{x+2}$

Once this is expanded algebraically, we get $x^2 - 10x - 4 = 0$.
Solving for $x$, we get two values of $x$. One is 10.38 and the other is -0.38
We ignore the negative value as it has no meaning. Thus the boat moved at a speed of 10.38 km/hour.