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 m2 by algebraic method, we have to assume the sides to be x m.

Then, side×side = area

⇒ xx = 625

x2 = 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 m2 and length is greater than breadth by 2, we have to assume breadth be x m and length be  +2 m,

Then, x(+2) = 63

x2 + 2x = 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 m2, to find base & altitude which are in ratio of 1:3, we assume base be x m & altitude be 3x m and the required equation will be

⋅x ⋅3x = 54

or, 3x2 = 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) (+1)2 = 2(x 3)

(ii) (+1)(x −2) = (x –1)(+3)

(iii) (+2)2 = 2x(x2 -1)

#### Solution

(i) (+1)2 = 2(x 3)

⇒ x+1 +2x =26

⇒ x+1 +22+6 = 0

⇒ x+7 = 0 (In standard form)

∴ Yes, it is a quadratic equation.

(ii) (+1)(2) = (1)(+3)

⇒ xx 2 =x2+2x 3

⇒ x2x x2x +3 = 0

⇒ 3x +1= 0 (In standard form)

∴ No, it is not a quadratic equation.

(iii) (+2)2 = 2x(x2 1)

⇒ x+4x +4 = 2x2x

⇒ x+4x +4 2x+2x = 0

⇒ 2x+x2 +6x +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(x2) and solutions / roots of the quadratic equation p(x2) = 0 are same.

In general, a real number α is called a root of the quadratic equation ax2 + bx + c = 0, a ≠ 0 if aα2 + 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 ax2 + bx + c = 0, a ≠ 0 if

2 + bα + c = 0 & aβ2 + 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.

ax2 + bx + c = 0

Let, ax2 + bx + c factorise to g(m+n)(p+q),

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

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

Now either g = 0 or (m+n) =0 or (p+q) = 0 or all of them can be zero.

From (m+n) = 0, we get

x n/m = α

From (p+q) = 0, we get

q/p = β

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

Let’s solve the following equation,  x2  3x  10 = 0

⇒ x2 + 2x 510 = 0

⇒ x(x + 2) 5(+ 2) = 0

⇒ (x + 2) (5) = 0

⇒ (x + 2) = 0 or (5) = 0

⇒ x = 2 or x = 5

∴ Roots of x2 3x 10 = 0 are 2 and 5

Verification:

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

Putting 2 in the equation,

(2)2  3(210 = 0

⇒ 4 +6 10 =0

⇒ 0 =0

Putting = 5 in the equation,

(5)2 3(510 = 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

ax2 + bx + c = 0, where (a ≠0)

If b = 0, then

ax2 + c = 0

x2 =  −c/a         …(i)

taking square roots on both side, we get

=  ±√−c/= ±√k  (let −c/= k )

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

e.g. 3x2+ (−27) = 0 , where a = 3, b = 0, c = −27

⇒ 3x−27 = 0 ⇒ 3x= 27

⇒ x27/3  ⇒ x= 9

⇒ √x= √9     ⇒ x= ±3

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

(x + m)2 − p = 0

(let, p = n2)

⇒ (x + m)2 − n2 = 0

⇒ (x + m)2 = n2       …(ii)

⇒ √(x + m)2 = √n2

x + m = ±n

⇒ x  = (−n+ m) or (n+ m)

e.g., (x+2)2−9 = 0

⇒ (x+2)2 = 9     …(iii)

⇒ (x+2)2 = 32

⇒ √(x+2)2 = √32  ⇒ x+2 = ±3

⇒ = −3 −2 or +3 −2

⇒ = −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 ax2 + bx + c = 0 can be transformed to the form (x + m)2 − n2 = 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 x2 + 4x − 5 = 0 using completing the square method.

Sol: x2 + 4x − 5 = 0

⇒ x2 + 2⋅ ⋅2 − 5 = 0

If we add 4 i.e., 2 to x2 + 2⋅ ⋅2, we can complete the square.

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

⇒ x2 + 2⋅ ⋅2+ 22 −5 = 0 + 22

⇒ (+2)2 − 5 = 0 + 4

⇒ (+2)2  = 4 + 5

⇒ (+2)2  = 9

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

Now, taking square on both side

⇒ √(+2)2  = √9

⇒ (x +2)  = ±3

⇒ x = −3 −2 or 3 −2

⇒ x = −5 or 1

Standard procedure to transform ax2 + bx + c = 0 to the form (x + m)2 − n2 = 0

↬ Divide each term by a, we get

x2 + b/a x + c/a  = 0/a

↬ Multiply and divide b/a x by 2, we get

x2 +2b/2a x + c/a  = 0/a

↬ Now we need to add (b/2a)to x2 + 2b/2a ⋅x to complete the square

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

x2 + 2b/2a x + (b/2a)c/a  = 0 + (b/2a)2

⇒ (+ b/2a)b2/4a2 − c/a

⇒ (b/2a)(b2 – 4ac)/4a2

↬ It is the required form, (x + m)2 = n2

where, m = b/2a and  n2 = (b2 – 4ac)/4a2

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

e.g., 3x2 − 5x + 2 = 0

⇒ 3/3x2 − 5x/3 + 2/3 = 0

⇒ x2 −2/2 ×5x/3 + 2/3 = 0

⇒ x2 −2⋅5/6⋅x + (-5/6)22/3 = 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

ax2 + bx + c = 0, where (a ≠0)

Dividing throughout by a, we get

x2 +bx/a + c/a = 0

⇒ (b/2a)2 − (b/2a)2 + c/a = 0

⇒ (b/2a)2 − (b2 − 4ac)/4a2 = 0

⇒ (b/2a)2 = (b2 − 4ac)/4a2

If b2 − 4ac ≥ 0, then by taking the square roots, we get

⇒ b/2a = ±√(b2 − 4ac)/2a

⇒ = −b ±√(b2 − 4ac)/2a

So, the roots of ax2 + bx + c = 0 are

−b −√(b2 − 4ac)/2a and −b +√(b2 − 4ac)/2a
Thus, if b2 – 4ac ≥ 0, then the roots of the quadratic equation ax2 + bx + c = 0 are given by

−b±√(b2−4ac)/2a

This is known as Quadratic Formula.

↪ If b2 − 4ac < 0, the equation will have no real roots.

EXAMPLE: Let’s find the the roots of 3x2 − 5x + 2 = 0 using quadratic formula.

Here, a = 3, b = −5, c = 2.

b2 − 4ac = (−5)2 − 4⋅3⋅2

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

∴ −b ±√(b2 − 4ac)/2a

⇒ = −(−5) ±√(−5)2 − 4⋅3⋅2/2⋅3

⇒ 5 ±√25 − 24/6

⇒ 5 ±1/6

⇒ 5−1/6  or 5+1/6

⇒ 4/6  or 6/6

⇒ = 2/or 1

Thus, the roots are 2/and 1.

## Nature of Roots

Roots of the quadratic equation ax2 + bx + c = 0 are given by

−b ±√(b2 − 4ac)/2a

↪ If b2 − 4ac > 0, we get two distinct real roots

−b−√(b2−4ac)/2a and −b+√(b2−4ac)/2a

↪ If b2 − 4ac = 0, then

x=−b±√(0/2a=−b/2a

i.e., both roots are real and equal

↪ If b2 – 4ac < 0, let b2 − 4ac = − γ, then √− γ would not exist as there is no real number whose square is negative number.Hence, no real roots.

### Discriminant

Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not, it is called discriminant of the equation. It is represented as δ or Δ.

A quadratic equation ax2 + bx + 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) 2x2 −3x + 5 = 0

(ii) 3x2 −4√3x + 4 = 0

#### Solution

(i) 2x2 −3x + 5 = 0

Δ = b2 − 4ac

= (−3)− 4⋅2⋅5

= 9− 40

= − 31

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

(ii) 3x2 −4√3x + 4 = 0

Δ = b2 − 4ac

= (−4√3)− 4⋅3⋅4

= 16⋅3− 48

= 48− 48 = 0

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

−b/2a −(−4√3)/2⋅3

= 4√3/2⋅3

= 2/√3 .