# Relations between the Roots and the Coefficients of a Quadratic Equation.

How to find the relation between roots and the **coefficients** of a **quadratic equation**.

Let us the quadratic equation of the general form

ax^{2}+bx+c=0 where **a** does not equal zero. **a** is a coefficient of **x ^{2}**,

**b**is a coefficient of x, and

**c**, is the constant term.

Now,

The standard form of a quadratic equation is

ax2+bx+c=0 where aâ‰ 0

Divided the equation by** a**

X^{2}+bx/a +c/a=0

Take constant terms to the R.H.S

X^{2}+bx/a=-c/a

To complete the square on the L.H.S add (b/2a)^{2} to both sides.

X^{2}+bx/a+ (b/2a)^{2}=(b/2a)^{2}-c/a

(x+b/2a)^{2}=b^{2}/4a^{2}-c/a

(x+b/2a)^{2}= b^{2}-4ac/4a^{2}

Square roots on both side

x+b/2a=Â±âˆš(b^2-4ac/2a

x=-b/2aÂ±âˆš(b^2-4ac/2a

x=(-bÂ±âˆš(b^2-4ac)/2a

Hence the solution of quadratic ax^{2}+bx+c=0 where aâ‰ 0 is given by

x=-bÂ±âˆš(b^2-4ac/2a

**Let Î± and Î² be the roots of ax ^{2}+bx+c=0 where aâ‰ 0 such that.**

**Î±=-b+âˆš(b^2-4ac/2a ****and Î²=-b-âˆš(b^2-4ac/2a**

**Therefore,**

Î±+Î²=(**-b+âˆš(b^2-4ac/2a**** ) + (-b-âˆš(b^2-4ac/2a)**

Î±+Î²=**-b+âˆš(b^2-4ac-**** b-âˆš(b^2-4ac/****2a**

Î±+Î²=-2b/2a

Î±+Î²=-b/a

**Sum of the roots =S=-b/a**

Sum of the roots =S=-coefficient of x/coefficient of x^{2}

**Again,**

**Î±Î²=(****-b+âˆš(b^2-4ac/2a****)( -b-âˆš(b^2-4ac/2a )**

**Î±Î² =(****-b+âˆš(b^2-4ac****)( -b-âˆš(b^2-4ac)/4a ^{2}**

**Î±Î²=**(-b)2-(**âˆš(b^2-4ac**)^{2}/4a^{2}

**Î±Î² =**b^{2}-b^{2}+4ac/4a^{2}

**Î±Î²=**4ac/4a.a

**Î±Î²= c/a**

Product of the roots =P=c/a

Product of the roots =P=constant term /coefficient of x^{2}

*The above results are helpful in expressing symmetric functions of the roots in terms of the coefficients of the quadratic equations.*

Therefore, Sum of the roots (Î±+Î² =-coefficient of x/coefficient of x^{2}) and the product of the roots (Î±Î²=constant term/coefficient of x^{2}) represent the required relations between roots (Î± andÎ²) and coefficients (a, b and c) of equation ax^{2}+bx+c=0

**For example:-** if the roots of the equation

5x^{2}-4x+7=0 be Î± and Î², than

Sum of roots =-coefficient of x/coefficient of x^{2}

Sum of roots =-(-4)/5

Sum of roots=4/5

Product of roots=constant term/coefficient of x^{2}

Product of roots=7/5

## Sum of Roots of a quadratic equation:-

The sum of the roots of a quadratic equation is equal to the negation of the coefficient of** x** divided by the coefficient of** x ^{2}**.it is denoted by

**S.**

Sum of roots =S=-b/a

Sum of roots= -coefficient of x/coefficient of x^{2}

**For example:- **Sum of roots of quadratic equation

X^{2}-5x+6=0

Compare the standard form of quadratic equation

ax2+bx+c=0 where aâ‰ 0

a=1

b=-5

c=6

Sum of roots =S=-b/a=-(-5)/1=5

## Product of Roots of a quadratic equation:-

The products of the roots of a quadratic equation are equal to the constant term divided by the coefficient of x^{2}.it is denoted by P.

**Product of roots =P=c/a**

Product of roots =constant term /coefficient of x^{2}

For example:-Product of roots of quadratic equation

2x^{2}+6x+3=0

Compare the standard form of quadratic equation

ax2+bx+c=0 where aâ‰ 0

a=2

b=6

c=3

Product of roots=c/a

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