Polynomial equations of second degree in one variable are called Quadratic equations.

(Polynomials are algebraic expressions in which powers of variables are whole numbers).

The general form of a quadratic equation is , where a, b, c are real numbers, a ≠ 0.

Ancient mathematicians encountered quadratic equations while solving area related problems. When algebraic equations were applied to find the sides of a square or a rectangle (or any geometrical shape) of known area, the equations so obtained were of second degree.

e.g., to find the sides of a square whose area is 625 m^{2} by algebraic method, we have to assume the sides to be *x* m.

Then, side×side = area

⇒ *x*⋅*x* = 625

⇒ *x*^{2} = 625

This is the required equation which must be solved to find the sides of the square.

This equation is a **polynomial equation** of second degree, therefore it is a quadratic equation.

Similarly, to find sides of a rectangle whose area is 63 m^{2} and length is greater than breadth by 2, we have to assume breadth be *x* m and length be *x *+2 m,

Then, *x*(*x *+2) = 63

*x*^{2} + 2*x* = 63

This is a required equation to be solved which is also a quadratic equation.

Similarly, to find sides of a triangle or radius of a circle for a known area, the required equations to be solved will be quadratic equations.

e.g., for a triangle of area 54 m^{2}, to find base & altitude which are in ratio of 1:3, we assume base be *x* m & altitude be 3*x* m and the required equation will be

⋅x* *⋅3*x* = 54

or, 3x^{2} = 108

- Some non-quadratic equations may appear to be quadratic equations. To find if an equation is quadratic or not, express it in standard form.

#### Try These

Check whether the following equations are quadratic equations.

(i) (*x *+1)^{2} = 2(*x* *−*3)

(ii) (*x *+1)(*x −*2) = (*x –*1)(*x *+3)

(iii) (*x *+2)^{2} = 2*x*(*x*^{2} -1)

#### Solution

(i) (*x *+1)^{2} = 2(*x* *−*3)

⇒ *x*^{2 }+1 +2*x *=2*x **−*6

⇒ *x*^{2 }+1 +2*x **−*2*x *+6 = 0

⇒ *x*^{2 }+7 = 0 (In standard form)

∴ Yes, it is a quadratic equation.

(ii) (*x *+1)(*x **−*2) = (*x **−*1)(*x *+3)

⇒ *x*^{2 }*−**x* *−*2 =*x*^{2}+2*x* *−*3

⇒ *x*^{2}*−**x* *−*2 *−**x*^{2 }*−*2*x* +3 = 0

⇒ *−*3*x* +1= 0 (In standard form)

∴ No, it is not a quadratic equation.

(iii) (*x *+2)^{2} = 2*x*(*x*^{2} *−*1)

⇒ *x*^{2 }+4*x* +4 *= *2*x*^{3 }*−*2*x*

⇒ *x*^{2 }+4*x* +4 *−*2*x*^{3 }+2*x = *0

⇒ *−*2*x*^{3 }+*x*^{2 }+6*x* +4= 0 (In standard form)

∴ No, it is not a quadratic equation.

# Solution / Root of a Quadratic Equations

A quadratic polynomial can have at most two zeroes which means that a quadratic equation can have at most two solutions.

Solutions of a quadratic equations are also known as the roots of the equation.

Zeroes of a quadratic polynomial *p(x ^{2}) *and solutions / roots of the quadratic equation

*p(x*are same.

^{2}) = 0In general, a real number α is called a root of the quadratic equation a*x*^{2} + b*x* + c = 0, a ≠ 0 if aα^{2} + bα + c = 0. We also say that *x* = α is a solution of the quadratic equation, or α satisfies the quadratic equation.

If a quadratic equation has two distinct roots, real numbers **α **& **β **are called roots of the quadratic equation a*x*^{2} + b*x* + c = 0, a ≠ 0 if

aα^{2} + bα + c = 0 & aβ^{2} + bβ + c = 0

### Solution of Quadratic Equations by Factorisation

In this method, we factorise the left hand expression of a quadratic equation to get two linear algebraic factors, then either of these factors or both would be equal to 0. Now, we equate each algebraic factors to 0 and solve them to get the required values of *x.*

a*x*^{2} + b*x* + c = 0

Let, a*x*^{2} + b*x* + c factorise to g**(m x +n)(px +q)**,

where g is constant factor and (m*x *+n) and (p*x *+q) are linear algebraic factors, then

g**(m x +n)(px +q)** = 0, where g, m, n , p and r are constants.

Now either g = 0 or **(m x +n)** =0 or

**(p**= 0 or all of them can be zero.

*x*+q)From (m*x *+n) = 0, we get

*x *= *−*n/m = α

From (p*x *+q) = 0, we get

*x *= *−*q/p = β

So, α & β are the solution of the equation.

Let’s solve the following equation, *x*^{2} *−* 3*x* *−* 10 = 0

⇒ *x*^{2} + 2*x* *−*5*x **−*10 = 0

⇒ *x*(*x* + 2) *−*5(*x *+ 2) = 0

⇒ **( x + 2) (x **

*−*

**5)**= 0

⇒ (*x* + 2) = 0 or (*x **−*5) = 0

⇒ *x* = *−*2 or *x* = 5

∴ Roots of *x*^{2} *−*3*x* *−*10 = 0 are *−*2 and 5

Verification:

We can verify the above equation for (*−*2,5)

Putting *x *= *−*2 in the equation,

(*−***2**)^{2} *−* 3(*−***2**) *−*10 = 0

⇒ 4 +6 *−*10 =0

⇒ 0 =0

Putting *x *= 5 in the equation,

(**5**)^{2} *−*3(**5**) *−*10 = 0

⇒ 25 *−*15 *−*10 =0

⇒ 0 = 0 Both *−*2 & 5 satisfies the given equation, therefore they are the roots of the equation.

### Solution of Quadratic Equations by Completing the Square

Let’s consider an equation in standard form

a*x*^{2} + b*x* + c = 0, where (a ≠0)

If b = 0, then

a*x*^{2} + c = 0

⇒ *x*^{2} = ^{−c}/_{a …}(i)

taking square roots on both side, we get

*x *= ±√^{−c}/_{a }= ±√k (let ^{−c}/_{a }= k )

If k >0, the equation will have real roots because negative numbers don’t have roots.

e.g. 3*x*^{2}+ (−27) = 0 , where a = 3, b = 0, c = −27

⇒ 3*x*^{2 }−27 = 0 ⇒ 3*x*^{2 }= 27

⇒ *x*^{2 }= ^{27}/_{3} ⇒ *x*^{2 }= 9

⇒ √*x*^{2 }= √9 ⇒ *x*= ±3

Let’s consider a quadratic equation in the form given below,

(*x* + m)^{2} − p = 0

(let, p = n^{2})

⇒ (*x* + m)^{2} − n^{2} = 0

⇒ (*x* + m)^{2} = n^{2 } …(ii)

⇒ √(*x* + m)^{2} = √n^{2 }

⇒ *x* + m = ±n

⇒ *x* = (−n+ m) or (n+ m)

e.g., (*x*+2)^{2}−9 = 0

⇒ (*x*+2)^{2} = 9 …(iii)

⇒ (*x*+2)^{2} = 3^{2}

⇒ √(*x*+2)^{2} = √3^{2} ⇒ *x*+2 = ±3

⇒ *x *= −3 −2 or +3 −2

⇒ *x *= −5 or 1

Roots of (*x*+2)^{2}−9 = 0 are −5 & 1.

In both the examples above, in equation (i) & (ii) ,the term containing *x* is completely inside a square and we found the roots easily by taking the square roots.

Quadratic equations in the form a*x*^{2} + b*x* + c = 0 can be transformed to the form (*x* + m)^{2} − n^{2} = 0, then it can easily be solved. This method is called **completing the square**.

This method is very helpful where quadratic equations can’t be solved through factorisation method.

We need to transform the given equation in such a way that expression in LHS become a complete square containing* x* and expression in RHS would contain constant only. We will need standard identities to complete the squares. Then we can take the square roots on both side and solved the equation.

*e.g.* Let’s find roots of *x*^{2} + 4*x* − 5 = 0 using completing the square method.

*Sol: x*^{2} + 4*x* − 5 = 0

⇒ *x*^{2} + 2⋅ *x *⋅2 − 5 = 0

If we add 4 i.e., 2^{2 } to *x*^{2} + 2⋅ *x *⋅2, we can complete the square.

Also, 4 must be added to other side of equality,

⇒ *x*^{2} + 2⋅ *x *⋅2+ 2^{2} −5 = 0 + 2^{2}

⇒ (*x *+2)^{2} − 5 = 0 + 4

⇒ (*x *+2)^{2} = 4 + 5

⇒ (*x *+2)^{2} = 9

[* It is same as equation (iii) in above examples.]

Now, taking square on both side

⇒ √(*x *+2)^{2} = √9

⇒ (*x* +2) = ±3

⇒ *x* = −3 −2 or 3 −2

⇒ *x* = −5 or 1

Standard procedure to transform a*x*^{2} + b*x* + c = 0 to the form (*x* + m)^{2} − n^{2} = 0

↬ Divide each term by a, we get

* x*^{2} + ^{b}/_{a} *x* + ^{c}/_{a} = ^{0}/_{a}

↬ Multiply and divide ^{b}/_{a} *x *by 2, we get

* x*^{2} +^{2b}/_{2a} *x* + ^{c}/_{a} = ^{0}/_{a}

↬ Now we need to add (^{b}/_{2a})^{2 }to *x*^{2} + ^{2b}/_{2a} ⋅*x *to complete the square

Add (^{b}/_{2a})^{2 }on both sides of the equation and complete the square

* x*^{2} + ^{2b}/_{2a} *x* + **( ^{b}/_{2a})^{2 }**+

^{c}/

_{a}= 0 +

**(**

^{b}/_{2a})^{2} ⇒ (*x *+ ^{b}/_{2a})^{2 }= ^{b2}/_{4a2}^{ −} ^{c}/_{a}

⇒ (*x *+ ^{b}/_{2a})^{2 }= ^{(b2 – 4ac)}/_{4a2}

↬ It is the required form, (*x* + m)^{2} = n^{2}

where, m = ^{b}/_{2a} and n^{2} = ^{(b2 – 4ac)}/_{4a2}

↬ Now take square on both sides ans solve the equation.

*e.g., *3*x*^{2} − 5*x* + 2 = 0

⇒ ^{3}/_{3}*x*^{2} − ^{5x}/_{3} + ^{2}/_{3} = 0

⇒ *x*^{2} −^{2}/_{2} ×^{5x}/_{3} + ^{2}/_{3} = 0

⇒ *x*^{2} −2⋅^{5}/_{6}*⋅x* + (^{-5}/_{6})^{2}+ ^{2}/_{3} = 0 + (^{-5}/_{6})^{2}

⇒ (*x* − ^{5}/_{6})^{2} = ^{25}/_{36}* −* ^{2}/_{3}

⇒ (*x* − ^{5}/_{6})^{2} = ^{(25-24)}/_{36}

⇒ √(*x* − ^{5}/_{6})^{2} = √^{1}/_{36}

⇒ (*x* − ^{5}/_{6}) = ±^{1}/_{6}

⇒ *x* = −^{1}/_{6} +^{5}/_{6 }or ^{1}/_{6} +^{5}/_{6}

⇒ *x* = ^{4}/_{6 }or ^{6}/_{6}

⇒ *x* = ^{2}/_{3 }or 1

## Quadratic Formula

a*x*^{2} + b*x* + c = 0, where (a ≠0)

Dividing throughout by a, we get

*x*^{2} +^{bx}/_{a} + ^{c}/_{a} = 0

⇒ (*x *+ ^{b}/_{2a})^{2} − (^{b}/_{2a})^{2} + ^{c}/_{a} = 0

⇒ (*x *+ ^{b}/_{2a})^{2} − ^{(b2 − 4ac)}/_{4a2} = 0

⇒ (*x *+ ^{b}/_{2a})^{2} = ^{(b2 − 4ac)}/_{4a2}

If b^{2} − 4ac ≥ 0, then by taking the square roots, we get

⇒ *x *+ ^{b}/_{2a} = ^{±√(b2 − 4ac)}/_{2a}

⇒ *x *= ^{−b ±√(b2 − 4ac)}/_{2a}

So, the roots of a*x*^{2} + b*x* + c = 0 are

^{−b −√(b2 − 4ac)}/_{2a }and ^{−b +√(b2 − 4ac)}/_{2a}

Thus, if b^{2} – 4ac ≥ 0, then the roots of the quadratic equation a*x*^{2} + b*x* + c = 0 are given by

^{−b±√(b2−4ac)}/_{2a}

This is known as **Quadratic Formula**.

↪ If b^{2} − 4ac < 0, the equation will have no real roots.

EXAMPLE: Let’s find the the roots of 3*x*^{2} − 5*x* + 2 = 0 using quadratic formula.

Here, a = 3, b = −5, c = 2.

b^{2} − 4ac = (−5)^{2} − 4⋅3⋅2

= 25 − 24 = 1 > 0, so the equation has real roots.

∴ *x *= ^{−b ±√(b2 − 4ac)}/_{2a}

⇒ *x *= ^{−(−5) ±√(−5)2 − 4⋅3⋅2}/_{2⋅3}

⇒ *x *= ^{5 ±√25 − 24}/_{6}

⇒ *x *= ^{5 ±1}/_{6}

⇒ *x *= ^{5−1}/_{6 }or ^{5+1}/_{6}

⇒ *x *= ^{4}/_{6 }or ^{6}/_{6}

⇒ *x *= ^{2}/_{3 }or 1

Thus, the roots are ^{2}/_{3 }and 1.

## Nature of Roots

Roots of the quadratic equation a*x*^{2} + b*x* + c = 0 are given by

*x *= ^{−b ±√(b2 − 4ac)}/_{2a}

↪ If **b**^{2}** − 4ac > 0**, we get two distinct real roots

^{−b−√(b2−4ac)}/_{2a }^{and −b+√(b2−4ac)}/_{2a}

↪ If **b**^{2}** − 4ac = 0**, then

*x*=^{−b±√(0}/_{2a}=^{−b}/_{2a}

i.e., both roots are real and equal

↪ If **b**^{2}** – 4ac < 0**, let b^{2} − 4ac = − γ, then √− γ would not exist as there is no real number whose square is negative number.Hence, no real roots.

### Discriminant

Since **b**^{2}** – 4ac** determines whether the quadratic equation a*x*^{2} + b*x* + c = 0 has real roots or not, it is called **discriminant** of the equation. It is represented as **δ** or **Δ**.

**A quadratic equation** **a****x**^{2}** +** **b****x****+ c = 0 has**

↪ two distinct real roots, if Δ > 0

↪ two equal real roots, if Δ = 0

↪ no real roots, if Δ < 0.

#### Try These

Find the nature of the roots of the following

(i) 2*x*^{2} −3*x* + 5 = 0

(ii) 3*x*^{2} −4√3*x* + 4 = 0

#### Solution

(i) 2*x*^{2} −3*x* + 5 = 0

Δ = b^{2} − 4ac

= (−3)^{2 }− 4⋅2⋅5

= 9− 40

= − 31

∵ Δ < 0, the given equation has no real roots.

(ii) 3*x*^{2} −4√3*x* + 4 = 0

Δ = b^{2} − 4ac

= (−4√3)^{2 }− 4⋅3⋅4

= 16⋅3− 48

= 48− 48 = 0

∵ Δ = 0, the given equation has two equal real roots.

*x *= ^{−b}/_{2a }= ^{−(−4√3)}/_{2⋅3}

= ^{4√3}/_{2⋅3}

= ^{2}/_{√3} .

⏪ Polynomials | Quadrilaterals ⏩ |