CBSE Class 10 Mathematics Quadratic Equations Assignment Set 18

Question. Find the solutions of the quadratic equation \(ax^2+bx+c=0, a \neq 0\) by the method of completing the square.
Answer: Given equation is: \[ax^2 + bx + c = 0\] Since \(a \neq 0\), dividing both sides by \(a\): \[x^2 + \frac{b}{a}x + \frac{c}{a} = 0\] Subtracting \(\frac{c}{a}\) from both sides: \[x^2 + \frac{b}{a}x = -\frac{c}{a}\] Adding \(\left(\frac{b}{2a}\right)^2\) to both sides to complete the square: \[x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}\] \[\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}\] \[\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}\] Taking square root on both sides: \[x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}\] \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Question. Prove that sum of roots of equation \(ax^2+bx+c=0, a \neq 0\) is \(\left(-\frac{b}{a}\right)\) and product is \(\left(\frac{c}{a}\right)\).
Answer: Let the roots of the quadratic equation \(ax^2 + bx + c = 0\) be \(\alpha\) and \(\beta\). From the quadratic formula, the roots are: \[\alpha = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \text{and} \quad \beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\] Sum of the roots: \[\alpha + \beta = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a}\] \[\alpha + \beta = \frac{-b + \sqrt{b^2 - 4ac} - b - \sqrt{b^2 - 4ac}}{2a}\] \[\alpha + \beta = \frac{-2b}{2a} = -\frac{b}{a}\] Product of the roots: \[\alpha \beta = \left(\frac{-b + \sqrt{b^2 - 4ac}}{2a}\right) \left(\frac{-b - \sqrt{b^2 - 4ac}}{2a}\right)\] \[\alpha \beta = \frac{(-b)^2 - (\sqrt{b^2 - 4ac})^2}{4a^2}\] \[\alpha \beta = \frac{b^2 - (b^2 - 4ac)}{4a^2}\] \[\alpha \beta = \frac{4ac}{4a^2} = \frac{c}{a}\] Hence proved.

Question. Solve the following equations by the method of completing the square. (3-7)
\(2x^2+5x-3=0\)
Answer: Given equation is: \[2x^2 + 5x - 3 = 0\] Dividing both sides by 2: \[x^2 + \frac{5}{2}x - \frac{3}{2} = 0\] \[x^2 + \frac{5}{2}x = \frac{3}{2}\] To complete the square, add \(\left(\frac{5}{4}\right)^2\) to both sides: \[x^2 + \frac{5}{2}x + \left(\frac{5}{4}\right)^2 = \frac{3}{2} + \frac{25}{16}\] \[\left(x + \frac{5}{4}\right)^2 = \frac{24 + 25}{16} = \frac{49}{16}\] Taking the square root on both sides: \[x + \frac{5}{4} = \pm \frac{7}{4}\] Case 1: \[x = -\frac{5}{4} + \frac{7}{4} = \frac{2}{4} = \frac{1}{2}\] Case 2: \[x = -\frac{5}{4} - \frac{7}{4} = -\frac{12}{4} = -3\] Thus, the solutions are \(\left(\frac{1}{2}, -3\right)\).

Question. \(4x^2 - 8x + 3 = 0\)
Answer: Given equation is: \[4x^2 - 8x + 3 = 0\] Dividing by 4: \[x^2 - 2x + \frac{3}{4} = 0\] \[x^2 - 2x = -\frac{3}{4}\] Add \(1\) to both sides to complete the square: \[x^2 - 2x + 1 = 1 - \frac{3}{4}\] \[(x - 1)^2 = \frac{1}{4}\] Taking the square root on both sides: \[x - 1 = \pm\frac{1}{2}\] Case 1: \[x = 1 + \frac{1}{2} = \frac{3}{2}\] Case 2: \[x = 1 - \frac{1}{2} = \frac{1}{2}\] Thus, the solutions are \(\left(\frac{1}{2}, \frac{3}{2}\right)\).

Question. \(2x^2 + x - 1 = 0\)
Answer: Given equation is: \[2x^2 + x - 1 = 0\] Dividing by 2: \[x^2 + \frac{1}{2}x - \frac{1}{2} = 0\] \[x^2 + \frac{1}{2}x = \frac{1}{2}\] Add \(\left(\frac{1}{4}\right)^2\) to both sides: \[x^2 + \frac{1}{2}x + \frac{1}{16} = \frac{1}{2} + \frac{1}{16}\] \[\left(x + \frac{1}{4}\right)^2 = \frac{9}{16}\] Taking the square root on both sides: \[x + \frac{1}{4} = \pm\frac{3}{4}\] Case 1: \[x = -\frac{1}{4} + \frac{3}{4} = \frac{2}{4} = \frac{1}{2}\] Case 2: \[x = -\frac{1}{4} - \frac{3}{4} = -\frac{4}{4} = -1\] Thus, the solutions are \(\left(\frac{1}{2}, -1\right)\).

Question. \(4x^2 - 4x + 1 = 0\)
Answer: Given equation is: \[4x^2 - 4x + 1 = 0\] Dividing by 4: \[x^2 - x + \frac{1}{4} = 0\] \[\left(x - \frac{1}{2}\right)^2 = 0\] Taking the square root: \[x - \frac{1}{2} = 0\] Thus, the equal roots are \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

Question. \(x^2 - x - 56 = 0\)
Answer: Given equation is: \[x^2 - x - 56 = 0\] \[x^2 - x = 56\] Add \(\left(\frac{1}{2}\right)^2\) to both sides: \[x^2 - x + \frac{1}{4} = 56 + \frac{1}{4}\] \[\left(x - \frac{1}{2}\right)^2 = \frac{225}{4}\] Taking the square root on both sides: \[x - \frac{1}{2} = \pm\frac{15}{2}\] Case 1: \[x = \frac{1}{2} + \frac{15}{2} = 8\] Case 2: \[x = \frac{1}{2} - \frac{15}{2} = -7\] Thus, the solutions are \((-7, 8)\).

Question. Form a quadratic equation whose roots are \(-1\) and \(2\)
Answer: Let the roots be \(\alpha = -1\) and \(\beta = 2\). Sum of roots: \[S = \alpha + \beta = -1 + 2 = 1\] Product of roots: \[P = \alpha\beta = -1 \times 2 = -2\] The quadratic equation is given by: \[x^2 - Sx + P = 0\] \[x^2 - x - 2 = 0\]

Question. Form a quadratic equation whose roots are \(\sqrt{3}\) and \(-\sqrt{3}\)
Answer: Let the roots be \(\alpha = \sqrt{3}\) and \(\beta = -\sqrt{3}\). Sum of roots: \[S = \alpha + \beta = \sqrt{3} + (-\sqrt{3}) = 0\] Product of roots: \[P = \alpha\beta = \sqrt{3} \times (-\sqrt{3}) = -3\] The quadratic equation is given by: \[x^2 - Sx + P = 0\] \[x^2 - 3 = 0\]

Question. Find sum of roots of equation \(2x^2 - x - 1 = 0\)
Answer: Given quadratic equation is \(2x^2 - x - 1 = 0\). Comparing with \(ax^2 + bx + c = 0\), we get: \(a = 2, b = -1, c = -1\) The sum of roots is given by \(\left(-\frac{b}{a}\right)\): \[\text{Sum of roots} = -\frac{-1}{2} = \frac{1}{2}\]