CBSE · Class 10 · Mathematics · Quadratic EquationsExplain the method of completing the square for solving a quadratic equation in detail. Also, use this method to find the roots of the quadratic equation $2x^2 - 7x + 3 = 0$.

Method of Completing the Square

\nThe method of completing the square is an algebraic technique used to solve quadratic equations of the general form $ax^2 + bx + c = 0$ (where $a \neq 0$). This method involves transforming the quadratic polynomial into a perfect square trinomial on one side of the equation, making it easier to solve for the variable $x$.

Step-by-Step Procedure:

1. Make the coefficient of $x^2$ unity: If the coefficient of $x^2$ is not 1, divide the entire equation by $a$ so that the coefficient becomes 1. The equation becomes $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$.
2. Shift the constant term: Move the constant term $\frac{c}{a}$ to the right-hand side of the equation: $x^2 + \frac{b}{a}x = -\frac{c}{a}$.
3. Add the square of half the coefficient of $x$: Take the coefficient of $x$, which is $\frac{b}{a}$, divide it by 2 to get $\frac{b}{2a}$, square it to get $\left(\frac{b}{2a}\right)^2$, and add this value to both sides of the equation. This creates a perfect square on the left side: $x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$.
4. Express as a squared binomial: Rewrite the left-hand side as the square of a binomial: $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$.
5. Solve for $x$: Take the square root on both sides and solve for $x$ to obtain the roots: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Application to the Given Equation

\nGiven quadratic equation: $$2x^2 - 7x + 3 = 0$$

Step 1: Divide the entire equation by 2 to make the coefficient of $x^2$ equal to 1. $$x^2 - \frac{7}{2}x + \frac{3}{2} = 0$$

Step 2: Shift the constant term to the right-hand side. $$x^2 - \frac{7}{2}x = -\frac{3}{2}$|

Step 3: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-\frac{7}{2}$. Half of this is $-\frac{7}{4}$, and its square is $\left(-\frac{7}{4}\right)^2 = \frac{49}{16}$. $$x^2 - \frac{7}{2}x + \frac{49}{16} = -\frac{3}{2} + \frac{49}{16}$$

Step 4: Simplify both sides.\nOn the left-hand side, write the expression as a complete square: $\left(x - \frac{7}{4}\right)^2$.\nOn the right-hand side, take the LCM: $$\frac{-24 + 49}{16} = \frac{25}{16}$$\nSo, $$\left(x - \frac{7}{4}\right)^2 = \frac{25}{16}$|

Step 5: Take the square root on both sides. $$x - \frac{7}{4} = \pm \sqrt{\frac{25}{16}}$$ $$x - \frac{7}{4} = \pm \frac{5}{4}$|

Step 6: Solve for $x$ by considering both positive and negative cases.

• Case 1 (Positive): $$x - \frac{7}{4} = \frac{5}{4}$$ $$x = \frac{5}{4} + \frac{7}{4} = \frac{12}{4} = 3$$

• Case 2 (Negative): $$x - \frac{7}{4} = -\frac{5}{4}$| $$x = -\frac{5}{4} + \frac{7}{4} = \frac{2}{4} = \frac{1}{2}$|

Conclusion: The roots of the given quadratic equation are $x = 3$ and $x = \frac{1}{2}$.