**T**he quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula:

(-b±√(b²-4ac))/(2a)

Where a, b, c , and a ≠ 0 is called a quadratic equation.

The numbers a,b,c are called the coefficients of this equation.

A root of the quadratic equation is the value of x which satisfies the equation, or

We can say if α is the root of a quadratic equation (i) then (x – α ) is a factor of ax²+bx+c=0

- A quadratic equation has two and only two roots.
- A quadratic equation cannot have more than two different roots.

Discrimens D= b^{2}-4ac

1. If D > 0, twp distinct real roots.

2. If D < 0, no real roots

3. If D = 0, Two equal real roots

E.g. Does equation 6x^{2} – x – 2 have real roots?

Comparing by the standard equation ax^{2} + bx + c = 0, we got the values of a,b,c as

a = 6

b = -1

c = -2

By putting the values in D = b2 – 4ac

⇒D = (-1)2 – 4 × 6 × (-2)

⇒D = 1 + 48

⇒D = 49 > 0

So equation has two real but unequal roots exist.

Formation of the quadratic equation:

Let an equation ax^{2} + bx + c = 0 has two real roots α and β

It is formed as

⇒ x^{2} – (sum of roots)x + (product of roots) = 0

⇒ x^{2} – (α + β) + αβ = 0

Hence

Sum of roots = α + β = -b / a

Product of roots = αβ = c / a

TIP- If equation has one root as P + √Q then the second roots will be P – √Q and vice versa.

Methods of solving quadratic equations

Finding the roots of the equation is called solving the equation, we do this by Factorisation method. Let’s study this method first

If ax^{2} + bx + c can be factorised as (x – α)(x – β), then ax^{2} + bx + c = 0 is equivalent to (x – α)(x – β) = 0

Thus x = α or β

E.g. Solve 5x^{2} + 11x + 6 = 0

By factorization method,

⇒5x^{2} + 11x + 6 = 0

⇒5x^{2} + 5x + 6x + 6 = 0

⇒5x(x + 1) + 6(x + 1) = 0

⇒(x + 1)(5x + 6) = 0

⇒(x + 1) = 0 Or (5x + 6) = 0

⇒x = -1 Or x = -6 / 5

E.g. Solve 16x¬2 – 53x + 35 = 0

By factorization method

⇒16x^{2} – 53x + 35 = 0

This task now becomes difficult and time-consuming

We have a new method for solving this type of heavy questions

Sridharacharya’s Method: Using the quadratic formula, write the equation in the standard form ax¬2 + bx + c = 0, then roots are

α = (-b + √(b2 – 4ac)) / 2a, β = (-b – √(b2 – 4ac)) / 2a

E.g. Solve 16x¬2 – 53x + 35 = 0

⇒ α = (-b + √(b2 – 4ac)) / 2a, β = (-b – √(b2 – 4ac)) / 2a

⇒ α = (-(-53) + √(53)2 – 4 × 16 × 35)) / (2 × 16), β = (-(-53) – √(53)2 – 4 × 16 × 35)) / (2 × 16)

⇒ α = (53 + √(2809 – 2240)) / 32, β = (53 – √(2809 – 2240)) / 32

⇒ α = (53 + √569) / 32, β = (53 – √569) / 32

⇒ α = (53 + 24) / 32, β = (53 – 24) / 32

⇒ α = 77 / 32, β = 29 / 32

E.g. Solve 5×2 + 11x + 6 = 0

⇒ α = (-b + √(b2 -4ac)) / 2a, β = (-b – √(b2 – 4ac)) / 2a

⇒ α = (-11 + √(11)2 – 4 × 5 × 6)) / (2 × 5), β = (-11 – √(11)2 – 4 × 5 × 6)) / (2 × 5)

⇒ α = (-11 + √(121 – 120)) / 10, β = (-11 – √(121 – 120)) / 10

⇒ α = (-10) / 10, β = (-12) / 10

⇒ α = -1, β = -6 / 5

E.g. Find the relation between x and y gave that

1. 2×2 + 19x + 42 = 0

2. 4y2 + 43y + 30 = 0

First, let’s try to find the root of 1st and 2nd.

By solving 1st

⇒ 2×2 + 19x + 42 = 0

⇒ α = (-b + √(b2 – 4ac)) / 2a, β = (-b – √(b2 – 4ac)) / 2a

⇒ α = (-19 + √(19)2 – 4 ×2 × 42)) / (2 × 2), β = (-19 – √(19)2 – 4 × 2 × 42)) / (2 × 2)

⇒ α = (-19 + √(19)2- 4 × 2 × 42)) / (2 × 2), β = (-19 – √(19)2- 4 × 2 × 42)) / (2 × 2)

⇒ α = (-19 + √25) / 4, β = (-19- √25) / 4

⇒ α = (-19 + 5) / 4, β = (-19 – 5) / 4

⇒ α = -3.5, β = -6

By solving 2nd

⇒ 4y2 + 43y + 30 = 0

⇒ α = (-43 + √(43)2 – 4 × 4 × 30)) / (2 × 4), β = (-43 – √(43)2 – 4 × 4 × 30)) / (2 × 4)

⇒ α = (-43 + √(43)2- 4 × 4 × 30)) / (2 × 4), β = (-43 – √(43)2 – 4 × 4 × 30)) / (2 × 4)

⇒ α = (-43 + √1369) / 8, β = (-43 – √1369) / 8

⇒ α = (-43 + 37) / 8, β = (-43 – 37) / 8

⇒ α = (-6) / 8, β = (-80) / 8

⇒ α = (-3) / 4, β = -10

So by comparing we saw there is no direct relation between them.

E.g. Find the relation between x and y when

1. (x – 12)2 = 0

2. y2 = 144

By solving 1st

⇒ (x – 12)2 = 0

⇒ x – 12 = 0

⇒ x = 12

By solving 2nd

⇒ y2 = 144

⇒ y = ± 12

So we can say x ≥ y

E.g. Quantity I. x2 – x – 6 = 0

Quantity II. y2 + 7y + 12 = 0

Find a relation between x and y

By solving Ist

⇒ x2 – x – 6 = 0

⇒ x2 – 3x + 2x – 6 = 0

⇒ x(x – 3) + 2(x – 3) = 0

⇒ (x + 2)(x – 3) = 0

⇒ x = -2,3

By solving 2^{nd}

⇒ y2 + 7y + 12 = 0

⇒ y2 + 4y + 3y + 12 = 0

⇒ y(y + 4) + 3(y + 4) = 0

⇒ (y + 3)(y + 4) = 0

⇒ y = -3, -4

Quantity I > Quantity II

E.g. 5x – 2y = 4

7x + 5y = 7

Find the relation between x and y

In the chapter of equations, we found three methods to solve the linear equations kindly refer to that, so let’s first try to solve both the equation by any one method

As 5x – 2y = 4

x = (4 + 2y) / 5

By putting the value of x in the second equation

⇒ 7((4 + 2y) / 5) + 5y = 7

⇒ 28 + 14y + 25y = 35

⇒ 39y = 7

⇒ y = 7 / 39

Putting the value of y in first

⇒ 5x – 2(7 / 39) = 4

⇒ 195x – 14 = 156

⇒39x = 14

⇒ x = 14 / 39

So we found x > y

It is the equation where two equations for two variables, let’s try to solve where there are 3 variable and 2 equation.

E.g. 9x + 2y – z = 7

15x – 2y + 3 / 5z = 4

As simply observing both the equation we can eliminate “y” factor from both the equations by adding it, So

Dividing both sides by 24

y – z / 60 = 7 / 12

So y > z

For solving the quadratic equations, we can use factorisation method after that we will find the value of roots and compare them

E.g. 2×2 – 7x + 6 = 0

y2 – 3y + 2 = 0

Find relation between x and y

Let’s first try to find the roots of both the equations by factorisation method

⇒ 2×2 – 7x + 6 = 0

⇒ 2×2 – 4x – 3x + 6 = 0

⇒ 2x(x – 2) – 3(x – 2) = 0

⇒ (x – 2)(2x – 3) = 0

⇒ x = 2 Or x = 3 / 2

Solving second equation

⇒ y2 – 3y + 2 = 0

⇒ y2 – 2y – y + 2 = 0

⇒ y(y – 2) – (y – 2) = 0

⇒ (y – 1)(y – 2) = 0

y = 1 Or y = 2

Say x = 1.5 and y = 2; then x

If x = 2 and y = 1; then x > y

No specific relation can be established between x & y

E.g. x2 – 7√7x + 84 = 0

y2 – 5√5 y + 30 = 0

Now the coefficient of x or y is rational in both the equations

But we will stick to our concept

Rationalizing first equation

⇒ x2 – 7√7x + 84 = 0

⇒ x2 – 4√7x – 3√7x + 84 = 0

⇒ x(x – 4√7) – 3√7(x – 4√7) = 0

⇒ (x – 3)(x – 4√7) = 0

x = 3√7 Or x = 4√7

Now the second equation’s turn

⇒ y2 – 5√5y + 30 = 0

⇒ y2 – 3√5y – 2√5y + 30 = 0

⇒ y(y – 3√5) – 2√5(y – 3√5)

⇒ (y – 2√5)(y – 3√5)

y = 2√5 Or y = 3√5

So x > y

Sometimes it becomes difficult to solve questions from the factorization method because that becomes tedious. Sridharacharya’s Method helps to find root but requires rigorous practice to do fast.

α = (-b + √(b2 – 4ac)) / 2a, β = (-b – √(b2 – 4ac)) / 2a

E.g. 2×2 – 7x + 6 = 0

y2 – 3y + 2 = 0

Let’s find the root for the first equation by putting values

⇒ 2×2 – 7x + 6 = 0

⇒ {-(-7) ± √((-7)2 – 4 × 2× 6)}/(2 × 2)

⇒ {7 ± √(49 – 48)} / 4

⇒ {7 ± 1} / 4

⇒ 2, 3 / 2

Now find the root of the second equation

y2 – 3y + 2 = 0

⇒ {-(-3) ± √(-3)2 – 4 × 1 × 2)} / (2 × 1)

⇒ {3 ± √(9 – 8)} / 2

⇒ {3 ± 1} / 2

⇒ 2, 1

Say x = 1.5 and y = 2; then x

Of x = 2 and y = 1; then x > y

No specific relation can be established between x & y

E.g. x2 – 7√7x + 84 = 0

y2 – 5√5y + 30 = 0

Let’s find the root for the first equation by putting values

⇒ x2 – 7√7x + 84 = 0

⇒ {-(-7√7) ± √((-7√7)2 – 4 × 1 × 84)} / (2 × 1)

⇒ {7√7 ± √(343 – 336)} / 2

⇒ (7√7 ± √7) / 2

⇒ 4√7, 3√7

Now find the root of the second equation

⇒ y2 – 5√5y + 30 = 0

⇒ {-(-5√5) ± √((-5√5)2 – 4 × 1 × 30)} / (2 × 1)

⇒ {5√5 ± √(125 – 120)} / 2

={5√5 ± √5} / 2

⇒ 3√5, 2√5

As by observing x > y

This method is a special case for the quadratic equation when signs of coefficients x and the numerical value of any one equation is positive.

E.g. x2 + 2x + 2 = 0

y2 – y + 5 = 0

As we know the coefficient of x is the negative of the sum of roots of equation {-(α + β)}, if the coefficient is positive it means the sum of roots is negative implies that both roots are negative.

Similarly, we know the coefficient of the numerical value is the product of roots if the sign is positive so roots can be negative.

Let’s see this for all cases;

Sr. no. Sign of (α+β) Sign of (αβ) Signs of the α, β Example

1. +ve +ve Both positive x2 – 2x + 2

2. -ve +ve Both negative x2 + 2x + 2

3. +ve -ve α is positive, β is negative x2 – 2x – 2

4. -ve -ve α is negative, β is positive x2 + 2x – 2

If we got two equations of 1st and 2nd case then we simply write an answer without calculating. Let’s see how to do this;

E.g. x2 + 2x + 2 = 0

y2 – y + 5 = 0

Comparing both equations by x2 – (α + β)x + (αβ) = 0 in 1st equation we got (α + β) is –ve and (αβ) is +ve, in 2nd equation we got (α + β) and (αβ) both +ve.

So according to table 1st have both negative roots and 2nd have both positive roots

so y > x without calculation

Let’s take one more example to clear this concept

E.g. 5y2 + 11y + 6 = 0

16×2 – 53x + 35 = 0

Comparing both equations by x2 – (α + β)x + (αβ) = 0 in 1st equation we got (α + β) is –ve and (αβ) is +ve, in 2nd equation we got (α + β) and (αβ) both +ve.

So according to table 1st have both negative roots and 2nd have both positive roots.

E.g. 5x – 2y = 4

7x + 5y = 7

Find the relation between x and y

In the chapter of equations, we found three methods to solve the linear equations kindly refer to that, so let’s first try to solve both the equation by any one method

As 5x – 2y = 4

x = (4 + 2y) / 5

By putting the value of x in the second equation

⇒ 7((4 + 2y) / 5) + 5y = 7

⇒ 28 + 14y + 25y = 35

⇒ 39y = 7

⇒ y = 7 / 39

Putting the value of y in first

⇒ 5x – 2(7 / 39) = 4

⇒ 195x – 14 = 156

⇒39x = 14

⇒ x = 14 / 39

So we found x > y

It is the equation where two equations for two variables, let’s try to solve where there are 3 variable and 2 equation.

E.g. 9x + 2y – z = 7

15x – 2y + 3 / 5z = 4

As simply observing both the equation we can eliminate “y” factor from both the equations by adding it

So

Dividing both sides by 24

y – z / 60 = 7 / 12

So y > z

For solving the quadratic equations, we can use factorisation method after that we will find the value of roots and compare them.

E.g. 2×2 – 7x + 6 = 0

y2 – 3y + 2 = 0

Find relation between x and y

Let’s first try to find the roots of both the equations by factorisation method

⇒ 2×2 – 7x + 6 = 0

⇒ 2×2 – 4x – 3x + 6 = 0

⇒ 2x(x – 2) – 3(x – 2) = 0

⇒ (x – 2) (2x – 3) = 0

⇒ x = 2 Or x = 3 / 2

Solving second equation

⇒ y2 – 3y + 2 = 0

⇒ y2 – 2y – y + 2 = 0

⇒ y(y – 2) – (y – 2) = 0

⇒ (y – 1) (y – 2) = 0

y = 1 Or y = 2

Say x = 1.5 and y = 2; then x

If x = 2 and y = 1; then x > y

No specific relation can be established between x & y

E.g. x2 – 7√7x + 84 = 0

y2 – 5√5 y + 30 = 0

Now the coefficient of x or y is rational in both the equations

But we will stick to our concept

Rationalizing first equation

⇒ x2 – 7√7x + 84 = 0

⇒ x2 – 4√7x – 3√7x + 84 = 0

⇒ x(x – 4√7) – 3√7(x – 4√7) = 0

⇒ (x – 3)(x – 4√7) = 0

x = 3√7 Or x = 4√7

Now the second equation’s turn

⇒ y2 – 5√5y + 30 = 0

⇒ y2 – 3√5y – 2√5y + 30 = 0

⇒ y(y – 3√5) – 2√5(y – 3√5)

⇒ (y – 2√5)(y – 3√5)

y = 2√5 Or y = 3√5

So x > y

Sometimes it becomes difficult to solve questions from the factorization method because that becomes tedious. Sridharacharya’s Method helps to find root but requires rigorous practice to do fast.

α = (-b + √(b2 – 4ac)) / 2a, β = (-b – √(b2 – 4ac)) / 2a

E.g. 2×2 – 7x + 6 = 0

y2 – 3y + 2 = 0

Let’s find the root for the first equation by putting values

⇒ 2×2 – 7x + 6 = 0

⇒ {-(-7) ± √((-7)2 – 4 × 2× 6)}/(2 × 2)

⇒ {7 ± √(49 – 48)} / 4

⇒ {7 ± 1} / 4

⇒ 2, 3 / 2

Now find the root of the second equation

y2 – 3y + 2 = 0

⇒ {-(-3) ± √(-3)2 – 4 × 1 × 2)} / (2 × 1)

⇒ {3 ± √(9 – 8)} / 2

⇒ {3 ± 1} / 2

⇒ 2, 1

Say x = 1.5 and y = 2; then x

Of x = 2 and y = 1; then x > y

No specific relation can be established between x & y

E.g. x2 – 7√7x + 84 = 0

y2 – 5√5y + 30 = 0

Let’s find the root for the first equation by putting values

⇒ x2 – 7√7x + 84 = 0

⇒ {-(-7√7) ± √((-7√7)2 – 4 × 1 × 84)} / (2 × 1)

⇒ {7√7 ± √(343 – 336)} / 2

⇒ (7√7 ± √7) / 2

⇒ 4√7, 3√7

Now find the root of the second equation

⇒ y2 – 5√5y + 30 = 0

⇒ {-(-5√5) ± √((-5√5)2 – 4 × 1 × 30)} / (2 × 1)

⇒ {5√5 ± √(125 – 120)} / 2

={5√5 ± √5} / 2

⇒ 3√5, 2√5

As by observing x > y

This method is a special case for the quadratic equation when signs of coefficients x and the numerical value of any one equation is positive.

E.g. x2 + 2x + 2 = 0

y2 – y + 5 = 0

As we know the coefficient of x is the negative of the sum of roots of equation {-(α + β)}, if the coefficient is positive it means the sum of roots is negative implies that both roots are negative.

Similarly, we know the coefficient of the numerical value is the product of roots if the sign is positive so roots can be negative.

Let’s see this for all cases;

Sr. no. Sign of (α+β) Sign of (αβ) Signs of the α, β Example

1. +ve +ve Both positive x2 – 2x + 2

2. -ve +ve Both negative x2 + 2x + 2

3. +ve -ve α is positive, β is negative x2 – 2x – 2

4. -ve -ve α is negative, β is positive x2 + 2x – 2

If we got two equations of 1st and 2nd case then we simply write an answer without calculating

Let’s see how to do this;

E.g. x2 + 2x + 2 = 0

y2 – y + 5 = 0

Comparing both equations by x2 – (α + β)x + (αβ) = 0 in 1st equation we got (α + β) is –ve and (αβ) is +ve, in 2nd equation we got (α + β) and (αβ) both +ve.

So according to table 1st have both negative roots and 2nd have both positive roots

so y > x without calculation

Let’s take one more example to clear this concept

E.g. 5y2 + 11y + 6 = 0

16×2 – 53x + 35 = 0

So according to table 1st have both negative roots and 2nd have both positive roots.