Revision:

A polynomial in one variable having degree 1 is called a linear polynomial, and having degree 2 is called a quadratic polynomial.

Activity:

Classify the following polynomials into linear polynomials and quadratic polynomials :
5x + 9, x2 + 3x – 5, 3x – 7, 3x2 – 5x, 5x2 The equation obtained by taking the value of a quadratic polynomial zero, is called a quadratic equation.
x2 + 3x – 5 = 0; 3x2 – 5x = 0; 5x2 = 0 are quadratic equations formed by taking the value of the quadratic polynomials, in the above box, zero. Activity:

x2 + 3x – 5, 3x2 – 5x, 5x2. Write the polynomials in the index form.
x2 + 3x – 5, 3x2 – 5x + 0, 5x2 + 0x + 0
Coefficients of x2 are 1, 3, and 5. These coefficients are non-zero.
Coefficients of x are 3, – 5 and 0 respectively.
Constants terms are – 5, 0 and 0 respectively. Here constant terms of second and third polynomials are zero.

Standard Form of the Quadratic Equation:

The equation involving one variable and having the maximum index of the variable 2 is called a quadratic equation.

1. The equation ax2 + bx + c = 0 is called the standard form of quadratic equation. Here, a, b and c are real numbers and a ≠ 0.
2. In the equation ax2 + bx + c = 0, if b = 0, then the equation becomes ax2 + c = 0. This is also a quadratic equation.
3. Similarly, if c = 0, then ax2 + bx = 0 is a quadratic equation.
4. If b = 0 and c = 0, then ax2 = 0 is a quadratic equation.
In the quadratic equation ax2 + bx + c = 0, the constants a, b, c are called the quadratic coefficient, the linear coefficient and the constant term respectively.

Activity: Complete the following table: Decide which of the following are quadratic equations.   The values of the variable which satisfy the given quadratic equation are called the roots of the quadratic equation.

Activity:

If x = 5 is a root of the equation kx2 – 14x – 5 = 0 then find the value of k by completing the following activity.

One of the roots of the equation kx2 – 14x – 5 = 0 Let’s Remember:

(1) ax2 + bx + c = 0 is the standard form of the quadratic equation, where a, b, care real numbers and a is non-zero.
(2) The values of the variable which satisfy the equation are called solutions or the roots of the equation.

Activity: Factorize the following polynomials: Following are the methods for solving quadratic equations:

• Factorisation method
• Completing square method
• Formula method.

Solution of a Quadratic Equation by Factorisation

• Write the given equation in the form ax2 + bx + c = 0.
• Find the two linear factors of the LHS of the equation.
• Equate each linear factor to zero.
• Solve each equation obtained in 3 and write the roots of the given equation. Study the following example:

Solve the quadratic equation x2 + 8x + 15 = 0 by factorisation method:

 Steps Quadratic Equation: x2 + 8x + 15 =0 1. Split the middle term 8x as 3x and 5x. [Because 3x × 5x = 15x2 = x2 × 15.] x2 + 3x + 5x + 15 = 0 2. Find the factors of LHS x(x +3) + 5(x +3) = 0 (x + 3)(x + 5) = 0 3. If the product of two  numbers is zero, then at least one of them must be zero. x + 3 = 0 or x + 5 = 0 4. Solve each linear  equation. x = – 3 or x = – 5 5. Write the answer. The roots of the equation are – 3, – 5.

Solution of a Quadratic Equation by Completing Square Method:

The quadratic equation of the type x2 + 6x +2 = 0 cannot be solved by the method of factorisation, because we cannot find two factors of 2 whose sum is 6.
In such a case, completing square method is used for solving quadratic equations.
For solving quadratic equation by this method proceed as follows:

 (1) Write the given equation in the form ax2 + bx + c = 0 x2 + 6x + 2 = 0 (2) Considering the first two terms on LHS, find the third suitable square term to make the polynomial a perfect square. Comparing x2 + 6x with x2 + 2xy, 2xy = 6x ∴ y = 3 ∴ y2 = 9 x2 + 6x + 9 is a perfect square polynomial. (3) Add the square term and subtract the same. x2 + 6x + 2 = 0 ∴ x2 + 6x + 9 – 9 + 2 = 0 (4) Write the square of the first three terms and the last two terms. (x + 3)2 – (√7)2 = 0 (5) Factorise and equate each factor to zero. (x + 3 + √7) (x + 3 – √7) = 0 x + 3 + √7 = 0 or x + 3 – √7 = 0 (6) Find the value of x. x = – 3 – √7 or x = – 3 + √7

– 3 – √7, – 3 + √7 are the roots of the quadratic equation.

Formula for Solving a Quadratic Equation

ax2 + bx + c. Divide the polynomial by a(∵a ≠ 0) to get x2 + $$\frac{b}{a}$$x + $$\frac{c}{a}$$.

Let us write the polynomial x2 + $$\frac{b}{a}$$x + $$\frac{c}{a}$$ in the form of difference of two square numbers. Now we can obtain roots or solutions of equation x2 + $$\frac{b}{a}$$x + $$\frac{c}{a}$$ = 0
which is equivalent to ax2 + bx + c = 0.
ax2 + bx + c = 0   Solve the quadratic equation x2 – x – 6 = 0 graphically.
x2 – x – 6 = 0
∴ x2 = x + 6.
Let, y = x2 = x + 6.
We draw the graphs of y = x2 and y = x + 6  The graph of y = x2 is a parabola.
The graphs of y = x2 and y = x + 6 intersect each other at (3, 9) and (-2, 4).
∴ x = 3 or x = – 2 is the solution of the given quadratic equation.

Activity 1:

Write the quadratic equation having sum of the roots 10 and the product of the roots 9. Activity 2:

What is the quadratic equation having the root α = 2 and β = 5? Let’s Remember:

1. If α and β are the roots of the quadratic equation ax2 + bx + c = 0, then  2. The nature of the roots of the quadratic equation ax2 + bx + c = 0 depends on the value of b2 – 4ac. b2 – 4ac is called the discriminant and is denoted by ∆(the Greek letter)

3. If ∆ = 0, the roots are real and equal.
If ∆ > 0, the roots are real and unequal.
If ∆ < 0, the roots are not real.

4. The quadratic equation whose roots are α and α is x2 – (α + β)x + αβ = 0.

Application of Quratic Equation:

Quadratic equations are useful to solve problems arising in our day-to-day life.
The method of solving problems consists of the following three steps:

Step 1: Convert the word problem, into symbolic language, i.e. form mathematical equation by Identifying the relationship existing in the problem.

Step 2: Solve the quadratic equation thus formed.

Step 3: Interpret the solution of the equation Into verbal language. The appropriate solution/solutions satisfying the given conditions is/are to be considered.