Quadratic Formula – Solution of quadratic equation using factorization

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

Consider the **quadratic equation** \( x^2~-~3x~+~2\)

\(1^2~-~3~+~2\)

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

\( 2^2~-~6~+~2\)

Here, 1 and 2 satisfies the equation \(x^2~-~3x~+~2\)

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 \( ax^2~+~bx~+~c\)

\( ax^2~+~bx~+~c\)

- 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; \( x^2~+~5x~+~6\)

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

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

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

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 \(\times\)

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

\(= 2x(x – 2) -1 (x – 2)\)

\(= (2x – 1)(x – 2)\)

Therefore, \(2x^2~-~5x~+~2\)

The values of x for which \( 2x^2~-~5x~+~2\)

If \( (2x ~-~ 1)~(x~ -~ 2)\)

\(2x ~-~ 1\)

\(x ~-~ 2\)

Therefore, \( \frac 12 \)

## Solution of the quadratic equation using quadratic formula

Consider the equation \( ax^2~+~bx~+~c\)

Dividing the equation by a gives,

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

By using method of completing the square, we get

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

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

\( \left( x~+~\frac{b}{2a}\right)^2 \)

Roots of the equation is found by taking the square root of RHS. For that \( b^2~-~4ac\)

When \( b^2~-~4ac\)

\( \left( x~+~\frac{b}{2a}\right) \)

\( x \)

\( x \)

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

The equation will not have real roots if \( b^2~-~4ac\)

(1) is formula to find roots of the quadratic equation \(ax^2~+~bx~+~c\)

Example: Find the roots of the equation\( x^2~-~5x~+~6\)

Comparing the equation with \( ax^2~+~bx~+~c\)

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

\(b^2~-~4ac\)

Roots of the equations are \(\frac {-b~+~\sqrt{b^2~-~4ac}}{2a}\)

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