CBSE Class 10 Mathematics Quadratic Equations Notes Set 06

Quadratic Equation: Introduction and Methods of Solving

Quadratic Equation

Any equation of the form \( p(x) = 0 \), where \( p(x) \) is a polynomial of degree 2, is known as a quadratic equation. For example, \( 3x^2 + 7x + 5 = 0 \) is a quadratic equation. Many real life situations can be represented by quadratic equations. An equation of the form \( ax^2 + bx + c = 0 \), where a, b, c are real numbers and \( a \ne 0 \), is known as the standard form of a quadratic equation.

Roots of a Quadratic Equation

A real number \( \alpha \) is a root of the quadratic equation \( ax^2 + bx + c = 0 \), \( a \ne 0 \), if \( a\alpha^2 + b\alpha + c = 0 \). So any real number which satisfies a given quadratic equation is called a root of the quadratic equation. In the chapter on polynomials we learnt that if \( x = \alpha \) satisfies a polynomial \( p(x) \), that is \( p(\alpha) = 0 \), then \( \alpha \) is a zero of \( p(x) \). This means that the zeros of the quadratic polynomial \( ax^2 + bx + c \) and the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are the same. A quadratic equation can have at most two real roots.

Note. The equation \( ax^2 + bx + c = 0 \) is a quadratic equation if \( a \ne 0 \) (two roots). It is a linear equation if \( a = 0, b \ne 0 \) (one root), a contradiction if \( a = b = 0, c \ne 0 \) (no root) and an identity if \( a = b = c = 0 \) (infinitely many roots). A quadratic equation cannot have more than two roots; if an equation of this form is satisfied by more than two values of x, it is satisfied by every value of x and so it is an identity.

Methods of Solving a Quadratic Equation

(i) By factorisation method.
Step 1. Factorise the constant term of the given quadratic equation.
Step 2. Express the coefficient of the middle term as the sum or difference of the factors obtained in Step 1. Clearly, the product of these two factors will be equal to the product of the coefficient of \( x^2 \) and the constant term.
Step 3. Split the middle term in two parts obtained in Step 2.
Step 4. Factorise the quadratic equation obtained in Step 3.

(ii) By completing the square method.
Step 1. Obtain the quadratic equation \( ax^2 + bx + c = 0 \), \( a \ne 0 \).
Step 2. Make the coefficient of \( x^2 \) unity, if it is not unity, i.e. obtain \( x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0 \).
Step 3. Shift the constant term \( \dfrac{c}{a} \) to the R.H.S. to get \( x^2 + \dfrac{b}{a}x = -\dfrac{c}{a} \).
Step 4. Add the square of half of the coefficient of x, i.e. \( \left(\dfrac{b}{2a}\right)^2 \), on both sides to obtain \( x^2 + 2\left(\dfrac{b}{2a}\right)x + \left(\dfrac{b}{2a}\right)^2 = \left(\dfrac{b}{2a}\right)^2 - \dfrac{c}{a} \).
Step 5. Write the L.H.S. as the perfect square of a binomial expression and simplify the R.H.S. to get \( \left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2 - 4ac}{4a^2} \).
Step 6. Take the square root of both sides to get \( x + \dfrac{b}{2a} = \pm\sqrt{\dfrac{b^2 - 4ac}{4a^2}} \).
Step 7. Obtain the values of x by shifting the constant term \( \dfrac{b}{2a} \) to the R.H.S.

(iii) By using the quadratic formula. Consider \( ax^2 + bx + c = 0 \), \( a \ne 0 \). Dividing throughout by a we get \( x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0 \), that is \( x^2 + \dfrac{b}{a}x = -\dfrac{c}{a} \). Completing the square,

\[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}. \]

When \( b^2 - 4ac \ge 0 \), \( x + \dfrac{b}{2a} = \pm\dfrac{\sqrt{b^2 - 4ac}}{2a} \), hence

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]

This is known as the quadratic formula or Shreedharacharya's formula for finding the roots of a quadratic equation. Hence, if \( b^2 - 4ac \ge 0 \), the roots of \( ax^2 + bx + c = 0 \) are \( \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \) and \( \dfrac{-b - \sqrt{b^2 - 4ac}}{2a} \).

Discriminant. For the quadratic equation \( ax^2 + bx + c = 0 \), the expression \( D = b^2 - 4ac \) is called its discriminant. In terms of D, the two roots are \( \alpha = \dfrac{-b + \sqrt{D}}{2a} \) and \( \beta = \dfrac{-b - \sqrt{D}}{2a} \).

Note. If the discriminant \( D = b^2 - 4ac < 0 \), then the quadratic equation \( ax^2 + bx + c = 0 \) has no real roots.

Solved Examples

Example 1. Which of the following are quadratic equations? (i) \( x^2 - 6x\sqrt{x} + 2 = 0 \)    (ii) \( x + \dfrac{1}{x} = 1 \)
Answer:
(i) \( x^2 - 6x\sqrt{x} + 2 = 0 \) is not a quadratic equation, as \( x^2 - 6x\sqrt{x} + 2 \) is not a polynomial (the power of x is \( \tfrac{3}{2} \)).
(ii) \( x + \dfrac{1}{x} = 1 \Rightarrow \dfrac{x^2 + 1}{x} = 1 \Rightarrow x^2 + 1 = x \Rightarrow x^2 - x + 1 = 0 \). Since \( x^2 - x + 1 \) is a polynomial of degree 2, \( x^2 - x + 1 = 0 \) is a quadratic equation.

Example 2. Determine whether the given values are solutions of the given equation or not. (i) \( 3x^2 + 2x - 1 = 0 \); \( x = \dfrac{1}{3} \)    (ii) \( a^2x^2 - 3abx + 2b^2 = 0 \); \( x = \dfrac{a}{b} \), \( x = \dfrac{b}{a} \)
Answer:
(i) Putting \( x = \dfrac{1}{3} \) in the L.H.S.: \( 3 \times \dfrac{1}{9} + \dfrac{2}{3} - 1 = \dfrac{1}{3} + \dfrac{2}{3} - 1 = 0 = \) R.H.S. So \( x = \dfrac{1}{3} \) is a solution.
(ii) Putting \( x = \dfrac{a}{b} \): L.H.S. \( = a^2 \cdot \dfrac{a^2}{b^2} - 3ab \cdot \dfrac{a}{b} + 2b^2 = \dfrac{a^4}{b^2} - 3a^2 + 2b^2 \ne 0 \) in general.
Putting \( x = \dfrac{b}{a} \): L.H.S. \( = a^2 \cdot \dfrac{b^2}{a^2} - 3ab \cdot \dfrac{b}{a} + 2b^2 = b^2 - 3b^2 + 2b^2 = 0 = \) R.H.S.
Hence \( x = \dfrac{a}{b} \) is not a solution (in general), but \( x = \dfrac{b}{a} \) is a solution of the given equation.

Example 3. Find the value of K for which the given value is a solution of the equation. (i) \( x^2 + Kx + 3 = 0 \), \( x = 1 \)    (ii) \( x^2 - x(a + b) + K = 0 \), \( x = a \)
Answer:
(i) Since \( x = 1 \) is a solution, \( 1 + K + 3 = 0 \), so \( K = -4 \).
(ii) Since \( x = a \) is a solution, \( a^2 - a(a + b) + K = 0 \Rightarrow -ab + K = 0 \), so \( K = ab \).

Example 4. If \( x = \dfrac{2}{3} \) and \( x = -3 \) are the roots of the equation \( ax^2 + 7x + b = 0 \), find the values of a and b.
Answer:
Since \( x = \dfrac{2}{3} \) is a root: \( a \cdot \dfrac{4}{9} + 7 \cdot \dfrac{2}{3} + b = 0 \Rightarrow 4a + 9b = -42 \) ... (i)
Since \( x = -3 \) is a root: \( 9a - 21 + b = 0 \Rightarrow 9a + b = 21 \) ... (ii)
Solving (i) and (ii): a = 3, b = -6.

Example 5. Find the solutions of the quadratic equation \( x^2 + 6x + 5 = 0 \).
Answer:
\( x^2 + 6x + 5 = x^2 + 5x + x + 5 = x(x + 5) + 1(x + 5) = (x + 5)(x + 1) \).
So \( (x + 5)(x + 1) = 0 \) gives \( x = -5 \) or \( x = -1 \).
Therefore \( x = -1, -5 \) are the required solutions.

Example 6. Solve: \( 25x^2 - 30x + 9 = 0 \).
Answer:
\( (5x)^2 - 2(5x)(3) + (3)^2 = 0 \Rightarrow (5x - 3)^2 = 0 \).
This gives \( x = \dfrac{3}{5}, \dfrac{3}{5} \), or simply \( x = \dfrac{3}{5} \) as the required solution.

Example 7. Solve \( 64x^2 - 625 = 0 \).
Answer:
\( (8x)^2 - (25)^2 = 0 \Rightarrow (8x + 25)(8x - 25) = 0 \), that is \( 8x + 25 = 0 \) or \( 8x - 25 = 0 \).
Thus \( x = -\dfrac{25}{8}, \dfrac{25}{8} \) are solutions of the given equation.

Example 8. Solve the quadratic equation \( 16x^2 - 24x = 0 \).
Answer:
The equation may be written as \( 8x(2x - 3) = 0 \). This gives \( x = 0 \) or \( x = \dfrac{3}{2} \).
\( x = 0, \dfrac{3}{2} \) are the required solutions.

Example 9. Solve the following quadratic equation by the factorisation method: \( x^2 - 2ax + a^2 - b^2 = 0 \).
Answer:
Here the factors of the constant term \( a^2 - b^2 \) are \( (a - b) \) and \( (a + b) \). Also the coefficient of the middle term is \( -2a = -[(a - b) + (a + b)] \).
\( x^2 - \{(a - b) + (a + b)\}x + (a - b)(a + b) = 0 \)
\( \Rightarrow x\{x - (a - b)\} - (a + b)\{x - (a - b)\} = 0 \)
\( \Rightarrow \{x - (a - b)\}\{x - (a + b)\} = 0 \)
So \( x = a - b \) or \( x = a + b \).

Example 10. Solve: \( x^2 + 6x + 8 = 0 \) by completing the square.
Answer:
\( x^2 + 6x = -8 \). Adding the square of half of the coefficient of x, \( (3)^2 \), on both sides: \( x^2 + 6x + 9 = -8 + 9 \).
\( (x + 3)^2 = 1 \Rightarrow x + 3 = \pm 1 \Rightarrow x + 3 = 1 \) or \( x + 3 = -1 \).
So \( x = -2 \) or \( x = -4 \) are the solutions.

Example 11. By using the method of completing the square, show that the equation \( 4x^2 + 3x + 5 = 0 \) has no real roots.
Answer:
Dividing by 4: \( x^2 + \dfrac{3}{4}x = -\dfrac{5}{4} \). Adding \( \left(\dfrac{3}{8}\right)^2 \) on both sides:
\( x^2 + \dfrac{3}{4}x + \left(\dfrac{3}{8}\right)^2 = \left(\dfrac{3}{8}\right)^2 - \dfrac{5}{4} \Rightarrow \left(x + \dfrac{3}{8}\right)^2 = \dfrac{9}{64} - \dfrac{5}{4} = -\dfrac{71}{64} \).
Clearly the R.H.S. is negative. But \( \left(x + \dfrac{3}{8}\right)^2 \) cannot be negative for any real value of x. Hence the given equation has no real roots.

Example 12. Solve the quadratic equation \( x^2 - 7x - 5 = 0 \) using the quadratic formula.
Answer:
Comparing with \( ax^2 + bx + c = 0 \): a = 1, b = -7, c = -5. \( D = b^2 - 4ac = 49 + 20 = 69 > 0 \), so the equation has two real roots.
\( x = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{7 \pm \sqrt{69}}{2} \).
So \( x = \dfrac{7 + \sqrt{69}}{2}, \dfrac{7 - \sqrt{69}}{2} \) are the required solutions.

Example 13. Solve the quadratic equation \( 9x^2 - 9(a + b)x + 2a^2 + 5ab + 2b^2 = 0 \).
Answer:
Here \( A = 9 \), \( B = -9(a + b) \), \( C = 2a^2 + 5ab + 2b^2 \).
\( B^2 - 4AC = 81(a + b)^2 - 36(2a^2 + 5ab + 2b^2) = 81a^2 + 81b^2 + 162ab - 72a^2 - 180ab - 72b^2 = 9a^2 + 9b^2 - 18ab = 9(a - b)^2 \).
\( x = \dfrac{9(a + b) \pm 3(a - b)}{18} \). Taking the positive sign: \( x = \dfrac{9a + 9b + 3a - 3b}{18} = \dfrac{2a + b}{3} \). Taking the negative sign: \( x = \dfrac{9a + 9b - 3a + 3b}{18} = \dfrac{a + 2b}{3} \).
So \( x = \dfrac{2a + b}{3}, \dfrac{a + 2b}{3} \).

Check Your Progress

1. Solve by factorisation: \( 9x^2 + 15x - 14 = 0 \).
Answer:
\( 9x^2 + 21x - 6x - 14 = 3x(3x + 7) - 2(3x + 7) = (3x + 7)(3x - 2) = 0 \).
\( x = -\dfrac{7}{3}, \dfrac{2}{3} \)

2. Solve by completing the square: \( x^2 - 2bx - 1 = -b^2 \).
Answer:
\( x^2 - 2bx + b^2 = 1 \Rightarrow (x - b)^2 = 1 \Rightarrow x - b = \pm 1 \).
\( x = b - 1, b + 1 \)

3. Solve using the quadratic formula: \( (a - b)^2x^2 - (a - b)^2x - ab = 0 \).
Answer:
\( D = (a - b)^4 + 4ab(a - b)^2 = (a - b)^2(a + b)^2 \), so \( x = \dfrac{(a - b)^2 \pm (a - b)(a + b)}{2(a - b)^2} = \dfrac{(a - b) \pm (a + b)}{2(a - b)} \) (for \( a \ne b \)).
\( x = \dfrac{a}{a - b}, \dfrac{-b}{a - b} \)