**Quadratic Formula – Solution of quadratic equation using factorization**

A quadratic equation is of the form of \( ax^2~+~bx~+~c~=~0, a~≠~0, a,b, ~and~ c\) are real numbers.

Consider the **quadratic equation** \( x^2~-~3x~+~2\) = 0; substituting x=1 in LHS of the equation gives,

\(1^2~-~3~+~2\) = 0 which is equal to RHS of the equation.

Similarly, substituting x = 2 in LHS of the equation also gives,

\( 2^2~-~6~+~2\) = 0 which is equal to RHS of the equation.

Here, 1 and 2 satisfies the equation \(x^2~-~3x~+~2\) =0. Therefore, those are known as the solution of quadratic equation or roots of the equation. This also means that numbers 1 and 2 are the zeros of the polynomial \( x^2~-~3x~+~2\) .

We know that, a second degree **polynomial** will have at most two zeros. Therefore a quadratic equation will have at most two roots.

In general, if α is a root of the quadratic equation \( ax^2~+~bx~+~c\) = 0, a ≠ 0; then, \(aα^2~+~bα~+~c\) = 0. We can also say that x = α is a solution of the quadratic equation or α satisfies the equation

\( ax^2~+~bx~+~c\) =0.

- Roots of the quadratic equation \( ax^2~+~bx~+~c\) =0 is same as zeros of the polynomial \( ax^2~+~bx~+~c\).

By splitting the middle term, we can factorise quadratic polynomial.

For example; \( x^2~+~5x~+~6\) can be written as \(x^2~+~2x~+~3x~+~6\) ,

\( x^2~+~2x~+~3x~+~6\) =\( x(x~ +~ 2)~ +~ 3(x~ +~ 2)\)

= \((x ~+~ 2)~(x~ +~ 3)\)

Roots of the quadratic equation can be found by factorising the polynomial and equate it with zero.

Consider the quadratic equation \( 2x^2~-~5x~+~2\) = 0;

which is of the form of \(ax^{2}+ bx +c = 0\)

Here, -5x can be broken down into two parts \(-4x – x\),

as the multiplication of these parts result in \(4x^{2}\)

which is equal to a \(\times\) c = \(2x^{2} \times 2\) =

\(4x^{2}\)Thus equation becomes: \(2x^{2} – 5x + 2 = 2x^{2} – 4x – x + 2\)

\(= 2x(x – 2) -1 (x – 2)\) \(= (2x – 1)(x – 2)\)Therefore, \(2x^2~-~5x~+~2\) = 0 is same as \((2x~ -~ 1) (x~ -~ 2)\)=0

The values of x for which \( 2x^2~-~5x~+~2\) = 0 is same as the values of x for which \((2x ~- ~1)~(x~ -~ 2)\) = 0.

If \( (2x ~-~ 1)~(x~ -~ 2)\) = \(0\) ; then, either \((2x ~-~ 1)\) = \(0~ or~ (x ~- ~2)\) = 0

\(2x ~-~ 1\) = 0 gives, \(2x \) = 1, \(x \)= \( \frac 12 \)

\(x ~-~ 2\) = 0 give \(sx\) = 2

Therefore, \( \frac 12 \) and 2 are the roots of the equation \( 2x^2~-~5x~+~2\) = 0.

**Solution of the quadratic equation using quadratic formula**

Consider the equation \( ax^2~+~bx~+~c\) = 0, a ≠ 0.

Dividing the equation by a gives,

\( x^2~+~ \frac ba x~+~ \frac ca \) = 0

By using method of completing the square, we get

\( \left( x~+~\frac{b}{2a}\right)^2~-~\left( \frac {b}{2a}\right)^2 ~+~\frac ca \) = 0

\( \left( x~+~\frac{b}{2a}\right)^2 ~-~ \frac {b^2~-~4ac}{4a^2}\) = 0

\( \left( x~+~\frac{b}{2a}\right)^2 \) = \( \frac {b^2~-~4ac}{4a^2}\)

Roots of the equation is found by taking the square root of RHS. For that \( b^2~-~4ac\) should be greater than or equal to zero.

When \( b^2~-~4ac\) ≥ 0,

\( \left( x~+~\frac{b}{2a}\right) \) = ± \( \frac {\sqrt{b^2~-~4ac}}{2a}\)

\( x \)= \( -~\frac{b}{2a}~±~\frac {\sqrt{b^2~-~4ac}}{2a}\)

\( x \) = \(\frac {-b~±~\sqrt{b^2~-~4ac}}{2a}\) —-(1)

Therefore roots of the equation are, \(\frac {-b~+~\sqrt{b^2~-~4ac}}{2a}\) and \(\frac {-b~-~\sqrt{b^2~-~4ac}}{2a}\)

Equation will not have real roots if \( b^2~-~4ac\) < 0, because square root is not defined for negative numbers in real number system.

(1) is formula to find roots of the quadratic equation \(ax^2~+~bx~+~c\) = 0, which is known as quadratic formula.

Example: Find the roots of the equation\( x^2~-~5x~+~6\) = 0 using quadratic formula.

Comparing the equation with \( ax^2~+~bx~+~c\) = 0 gives,

a = 1, b = -5 and c = 6

\(b^2~-~4ac\) = \((-5)^2~-~4~×~1~×~6\) = 1

Roots of the equations are \(\frac {-b~+~\sqrt{b^2~-~4ac}}{2a}\) = \( \frac {5~+~1}{2} \) = \( \frac {6}{2}\) = 3 and \(\frac {-b~-~\sqrt{b^2~-~4ac}}{2a}\) = \( \frac {5~-~1}{2} \) = \( \frac {4}{2}\) = 2

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