# How To Derive Quadratic Equations From Their Roots (Step-By-Step)

If you know how to solve quadratic equation, that is not the end of the thing you should know. Now, do you know how to derive quadratic equations from their given roots? Don’t worry, okay? Because will show you everything today.

In today’s article, we will teach you how to construct or make quadratic equation from it given roots.

The roots of a quadratic equation are the values which you get after the quadratic equation has been correctly solved.

Sometimes the roots of a particular quadratic may be given and you can be asked to find its equation. Whenever you are being asked like that, then there are two methods you can use to solve that. You don’t need to panic since you are here.

## Methods To Derive Quadratic Equations From Their Given Roots

The following are an examples and methods of constructing quadratic equation from its given roots.

Method i:

In this first method, you would need to write the given roots of quadratic equation in the formula provided above

Note: Roots of quadratic equation are usually two.

That is, X-(first given root) X-(second given roots)=0.

Method ii:

The second method for constructing quadratic equation from its given roots involves the use of formula also. And the formula is:

X²-(sum of given roots)X+(product of given roots)=0

I.e, X²-(first given root+second given roots) X+(first given root×second given roots) =0

### Examples Of How To Construct Quadratic Equation From It Given Roots

The following are used as an illustrations of how to derive quadratic equation from its given roots.  Please read them carefully.

Below are the examples of how to derive quadratic equations from their roots: Example 1: -2/5 and -3/7

Solution

X-(-2/5)x-(-3/7

X+(2/5)x+(2/5)=0

pairing:

x²+(x+3/7)+2/5(x+3/7)

x²+3x/7+2x/5+6/35

LCM=35, then divide by 35

x²×35/1+3x×35/7+2x×35/5+6×35/35=0

35x²+15x+14x+6=0

35x²+29x+6=0

Example 2: -3 and +6

Solution

Let’s us method ii X²-(sum of given roots) X+(product of given roots)=0

X²-(-3+(+)6)x+(-3×(+6)>=0

X²+(3+6)x+(-18)=0

X²+3x+6x-18=0

X²+9x-18=0

Example 3: √3 and √5

Solution

X-(√3)x-(√5)=0

Pairing and grouping:

X(X-√5)-√3(X-√5)=0

X²-√5X-√3X+√15=0

Example 4: 4 and 3/2

Solution

X²-(4+3/2)x+(4×3/2)=0

X²-(11/2)x+6/1=0

LCM=2, then let divide them by 2

2X²-11x+12=0.

Conclusion

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