CBSE Class 10 Mathematics Quadratic Equations Notes Set 04

Completing the Square

Completing the square is a method used to solve quadratic equations by adding constants to both sides of the equation so that the left side becomes a perfect square trinomial.

For e.g. the number that should be added to the equation \( x^2 - 4x = 0 \) in order to make it a perfect square trinomial is 4 i.e. \( x^2 - 4x + 4 = 4 \Rightarrow (x - 2)^2 = 4 \) (\( \because a^2 - 2ab + b^2 = (a - b)^2 \)).

Algorithm:
Step (i) Let the quadratic equation be \( ax^2 + bx + c = 0,\ a \neq 0 \).
Step (ii) If coefficient of \( x^2 \) is not unity, make it unity i.e., obtain \( x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \).
Step (iii) Shift the constant term \( \frac{c}{a} \) on R.H.S. to get \( x^2 + \frac{b}{a}x = -\frac{c}{a} \).
Step (iv) Take the coefficient of x, divide it by 2 and square whatever you get and add this to both sides of the equation i.e. \( \left(\frac{b}{2a}\right)^2 \) on both sides to obtain \( x^2 + 2\left(\frac{b}{2a}\right)x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a} \).
Step (v) Write L.H.S. as the perfect square of a binomial expression and simplify R.H.S. to get \( \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \).
Step (vi) Take square root of both sides to get \( x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} \).
Step (vii) Obtain the values of x by shifting the constant term \( \frac{b}{2a} \) on RHS.

Example 1. Find the roots of the quadratic equation \( 2x^2 - 7x + 3 = 0 \) by the method of completing the square.
Answer:
The given quadratic equation is of the form \( ax^2 + bx + c = 0 \). As coefficient of \( x^2 \) is not unity, first we make it unity: \( x^2 - \frac{7}{2}x + \frac{3}{2} = 0 \).
Next we shift the constant term on R.H.S.: \( x^2 - \frac{7}{2}x = -\frac{3}{2} \).
Now add half of the square of the coefficient of x to both the sides: \( x^2 - 2 \times x \times \frac{7}{4} + \left(\frac{7}{4}\right)^2 = -\frac{3}{2} + \left(\frac{7}{4}\right)^2 \).
Next write L.H.S. as a perfect square of the binomial expression: \( \left(x - \frac{7}{4}\right)^2 = \frac{-24 + 49}{16} = \frac{25}{16} \).
Taking square root of both the sides: \( x - \frac{7}{4} = \pm\frac{5}{4} \Rightarrow x = \frac{7}{4} + \frac{5}{4} \) or \( x = \frac{7}{4} - \frac{5}{4} \Rightarrow x = 3 \) or \( x = \frac{1}{2} \).

Example 2. Find the roots of the quadratic equation \( 4x^2 + 4\sqrt{3}x + 3 = 0 \) by the method of completing the square.
Answer:
\( 4x^2 + 4\sqrt{3}x + 3 = 0 \Rightarrow x^2 + \sqrt{3}x + \frac{3}{4} = 0 \Rightarrow x^2 + \sqrt{3}x = -\frac{3}{4} \).
\( x^2 + 2 \times x \times \frac{\sqrt{3}}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 = \left(\frac{\sqrt{3}}{2}\right)^2 - \frac{3}{4} \Rightarrow \left(x + \frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} - \frac{3}{4} = 0 \).
\( \therefore \) Roots are \( \frac{-\sqrt{3}}{2} \) and \( \frac{-\sqrt{3}}{2} \).

Example 3. Find the roots of the quadratic equation \( 2x^2 + x + 4 = 0 \) by the method of completing the square.
Answer:
\( 2x^2 + x + 4 = 0 \Rightarrow x^2 + \frac{x}{2} + 2 = 0 \Rightarrow x^2 + 2 \times x \times \frac{1}{4} + \left(\frac{1}{4}\right)^2 = -2 + \left(\frac{1}{4}\right)^2 \).
\( \left(x + \frac{1}{4}\right)^2 = -2 + \frac{1}{16} = -\frac{31}{16} \).
Real roots of the given quadratic equation do not exist as the square of a real number cannot be negative.

Quadratic Formula (Sridharacharya Formula)

Sridharacharya was an Indian mathematician, Sanskrit pandit and a philosopher. He gave the quadratic formula to solve the quadratic equation, be it adfected or pure.\[ x = \frac{-b \pm \sqrt{D}}{2a} \]

Algorithm:
Step (i) Find the value of a, b and c by comparing the given quadratic equation with general quadratic equation i.e., \( ax^2 + bx + c = 0 \).
Step (ii) Find the discriminant of the quadratic equation, using \( D = b^2 - 4ac \).
Step (iii) Now find the roots of the equation, using \( x = \frac{-b + \sqrt{D}}{2a},\ \frac{-b - \sqrt{D}}{2a} \).

Example 4. Solve the following quadratic equations by using quadratic formula: (i) \( x^2 - 7x + 12 = 0 \); (ii) \( 2y^2 + 3y - 35 = 0 \).
Answer:
(i) Comparing the given equation \( x^2 - 7x + 12 = 0 \) with standard quadratic equation \( ax^2 + bx + c = 0 \), we get \( a = 1,\ b = -7 \) and \( c = 12 \). \( \therefore D = b^2 - 4ac = (-7)^2 - 4 \times 1 \times 12 = 1 \) and \( \sqrt{b^2 - 4ac} = \sqrt{1} = 1 \).
\( \because x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \Rightarrow x = \frac{7 \pm 1}{2 \times 1} = \frac{7 + 1}{2} \) or \( \frac{7 - 1}{2} = 4 \) or 3.
(ii) Comparing the given equation \( 2y^2 + 3y - 35 = 0 \) with equation \( ay^2 + by + c = 0 \) we get: \( a = 2,\ b = 3 \) and \( c = -35 \). \( \therefore D = b^2 - 4ac = (3)^2 - 4(2)(-35) = 289 \Rightarrow \sqrt{b^2 - 4ac} = \sqrt{289} = 17 \).
Hence, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm 17}{2(2)} \Rightarrow y = -5 \) or \( \frac{7}{2} \).

Example 5. Using the quadratic formula, solve the equation: \( a^2b^2x^2 - (4b^4 - 3a^4)x - 12a^2b^2 = 0 \).
Answer:
Comparing the given equation with \( Ax^2 + Bx + C = 0 \), we get: \( A = a^2b^2,\ B = -(4b^4 - 3a^4) \) and \( C = -12a^2b^2 \).
\( \therefore B^2 - 4AC = [-(4b^4 - 3a^4)]^2 - 4 \times a^2b^2 \times (-12a^2b^2) = 16b^8 + 9a^8 - 24a^4b^4 + 48a^4b^4 = 16b^8 + 9a^8 + 24a^4b^4 = (4b^4 + 3a^4)^2 \).
\( \sqrt{B^2 - 4AC} = 4b^4 + 3a^4 \). \( \therefore x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{(4b^4 - 3a^4) \pm (4b^4 + 3a^4)}{2 \times a^2b^2} \).
\( x = \frac{4b^4 - 3a^4 + 4b^4 + 3a^4}{2a^2b^2} = \frac{8b^4}{2a^2b^2} = \frac{4b^2}{a^2} \) or \( x = \frac{4b^4 - 3a^4 - 4b^4 - 3a^4}{2a^2b^2} = \frac{-6a^4}{2a^2b^2} = \frac{-3a^2}{b^2} \).

Example 6. For a quadratic equation \( ax^2 + bx + c = 0 \), where \( a \neq 0 \), prove that \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) (proof of quadratic formula).
Answer:
Given, \( ax^2 + bx + c = 0 \Rightarrow x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \Rightarrow x^2 + \frac{b}{a}x = -\frac{c}{a} \).
Adding \( \left(\frac{b}{2a}\right)^2 \) on both sides: \( x^2 + 2 \times \frac{b}{2a} \times x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 \Rightarrow \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \).
Taking square roots: \( x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} \Rightarrow x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \).
Hence \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).