Class 11: Quadratic Equations – Exercise 14.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Solve the following quadratic equations by factorization method ( 1 – 9).

Question 1: $\displaystyle x^2 +1 = 0$

Answer:

$\displaystyle x^2 +1 = 0$

$\displaystyle \Rightarrow x^2 - i^2 = 0$

$\displaystyle \Rightarrow (x-i)(x+i) = 0$

$\displaystyle \Rightarrow x = i \text{ or } x = -i$

$\displaystyle \text{Hence the root of the equation } x^2 +1 = 0 \text{ are } i \ \text{ and } \ (-i)$

$\displaystyle \\$

Question 2: $\displaystyle 9x^2 + 4 = 0$

Answer:

$\displaystyle 9x^2 + 4 = 0$

$\displaystyle \Rightarrow (3x)^2 - (2i)^2 = 0$

$\displaystyle \Rightarrow (3x-2i)(3x+2i) = 0$

$\displaystyle \Rightarrow x = \frac{2i}{3} \text{ or } x = -\frac{2i}{3}$

$\displaystyle \text{Hence the root of the equation } 9x^2 + 4 = 0 \text{ are } \frac{2i}{3} \ \text{ and } \ -\frac{2i}{3}$

$\displaystyle \\$

Question 3: $\displaystyle x^2 + 2x + 5 = 0$

Answer:

$\displaystyle x^2 + 2x + 5 = 0$

$\displaystyle x^2 + 2x + 1+4 = 0$

$\displaystyle \Rightarrow (x+1)^2 - (2i)^2 = 0$

$\displaystyle \Rightarrow (x+1-2i)(x+1+2i) = 0$

$\displaystyle \Rightarrow x = (-1+2i) \text{ or } x = (-1-2i)$

$\displaystyle \text{Hence the root of the equation } x^2 + 2x + 5 = 0 \text{ are } (-1+2i) \ \text{ and } \ (-1-2i)$

$\displaystyle \\$

Question 4: $\displaystyle 4x^2 - 12 x + 25 = 0$

Answer:

$\displaystyle 4x^2 - 12 x + 25 = 0$

$\displaystyle 4x^2 - 12 x + 9 + 16 = 0$

$\displaystyle \Rightarrow (2x-3)^2 - (4i)^2 = 0$

$\displaystyle \Rightarrow (2x-3-4i)(2x-3+4i) = 0$

$\displaystyle \Rightarrow x = \frac{3+4i}{2} \text{ or } x = \frac{3-4i}{2}$

$\displaystyle \Rightarrow x = (\frac{3}{2}+ 2i) \text{ or } x = (\frac{3}{2} - 2i)$

$\displaystyle \text{Hence the root of the equation } 4x^2 - 12 x + 25 = 0 \text{ are } (\frac{3}{2}+ 2i) \ \text{ and } \ (\frac{3}{2} - 2i)$

$\displaystyle \\$

Question 5: $\displaystyle x^2 + x + 1 = 0$

Answer:

$\displaystyle x^2 + x + 1 = 0$

$\displaystyle x^2 + x + \frac{1}{4} + \frac{3}{4} = 0$

$\displaystyle \Rightarrow (x+ \frac{1}{2})^2 - (\frac{\sqrt{3}}{2}i)^2 = 0$

$\displaystyle \Rightarrow (x+ \frac{1}{2} - \frac{\sqrt{3}}{2}i )(x+ \frac{1}{2} + \frac{\sqrt{3}}{2}i ) = 0$

$\displaystyle \Rightarrow x = (-\frac{1}{2}+ \frac{\sqrt{3}}{2}i) \text{ or } x = (-\frac{1}{2}- \frac{\sqrt{3}}{2}i)$

$\displaystyle \text{Hence the root of the equation } x^2 + x + 1 = 0 \text{ are } (-\frac{1}{2}+ \frac{\sqrt{3}}{2}i) \ \text{ and } \ (-\frac{1}{2}- \frac{\sqrt{3}}{2}i)$

$\displaystyle \\$

Question 6: $\displaystyle 4x^2 + 1 = 0$

Answer:

$\displaystyle 4x^2 + 1 = 0$

$\displaystyle \Rightarrow (2x)^2 - i^2 = 0$

$\displaystyle \Rightarrow (2x-i)(2x+i) = 0$

$\displaystyle \Rightarrow x = \frac{i}{2} \text{ or } x = \frac{-i}{2}$

$\displaystyle \text{Hence the root of the equation } 4x^2 + 1 = 0 \text{ are } \frac{i}{2} \ \text{ and } \ \frac{-i}{2}$

$\displaystyle \\$

Question 7: $\displaystyle x^2 - 4x + 7 = 0$

Answer:

$\displaystyle x^2 - 4x + 7 = 0$

$\displaystyle \Rightarrow x^2 - 4x + 4 + 3 = 0$

$\displaystyle \Rightarrow (x-2)^2 - (\sqrt{3}i)^2 = 0$

$\displaystyle \Rightarrow (x-2-\sqrt{3} i)(x-2+\sqrt{3}i) = 0$

$\displaystyle \Rightarrow x = (2+\sqrt{3} i) \text{ or } x = (2 - \sqrt{3} i)$

$\displaystyle \text{Hence the root of the equation } x^2 - 4x + 7 = 0 \text{ are } (2+\sqrt{3} i) \ \text{ and } \ (2 - \sqrt{3} i)$

$\displaystyle \\$

Question 8: $\displaystyle x^2 + 2x + 2 = 0$

Answer:

$\displaystyle x^2 + 2x + 2 = 0$

$\displaystyle \Rightarrow x^2 + 2x + 1+1 = 0$

$\displaystyle \Rightarrow (x+1)^2 - (i)^2 = 0$

$\displaystyle \Rightarrow (x+1-i)(x+1+i) = 0$

$\displaystyle \Rightarrow x = (-1+i) \text{ or } x = (-1-i)$

$\displaystyle \text{Hence the root of the equation } x^2 + 2x + 2 = 0 \text{ are } (-1+i) \ \text{ and } \ (-1-i)$

$\displaystyle \\$

Question 9: $\displaystyle 5x^2 - 6x+2 = 0$

Answer:

$\displaystyle 5x^2 - 6x+2 = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 5, \ b= -6 \text{ and } c = 2$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{6 + \sqrt{36 - 4 \times 5 \times 2} }{2 \times 5} \text{ and } \beta = \frac{6 - \sqrt{36 - 4 \times 5 \times 2} }{2 \times 5}$

$\displaystyle \Rightarrow \alpha = \frac{6 + \sqrt{-4} }{10} \text{ and } \beta = \frac{6 - \sqrt{-4} }{10}$

$\displaystyle \Rightarrow \alpha = \frac{6 + 2i }{10} \text{ and } \beta = \frac{6 - 2i }{10}$

$\displaystyle \Rightarrow \alpha = \frac{3}{5} + \frac{1}{5} i \text{ and } \beta = \frac{3}{5} + \frac{1}{5} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{3}{5} \pm \frac{1}{5} i$

$\displaystyle \\$

Solve the following quadratics:

Question 10: $\displaystyle 21x^2 + 9x + 1 = 0$

Answer:

$\displaystyle 21x^2 + 9x + 1 = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 21, \ b= 9 \text{ and } c = 1$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{-9 + \sqrt{81 - 4 \times 21 \times 1} }{2 \times 21} \text{ and } \beta = \frac{-9 - \sqrt{81 - 4 \times 21 \times 1} }{2 \times 21}$

$\displaystyle \Rightarrow \alpha = \frac{-9 + \sqrt{-3} }{42} \text{ and } \beta = \frac{-9 - \sqrt{-3} }{42}$

$\displaystyle \Rightarrow \alpha = \frac{-9 + \sqrt{3}i }{42} \text{ and } \beta = \frac{-9 - \sqrt{3}i }{42}$

$\displaystyle \Rightarrow \alpha = \frac{-3}{14} + \frac{\sqrt{3}}{42} i \text{ and } \beta = \frac{-3}{14} - \frac{\sqrt{3}}{42} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{-3}{14} \pm \frac{\sqrt{3}}{42} i$

$\displaystyle \\$

Question 11: $\displaystyle x^2 - x + 1=0$

Answer:

$\displaystyle x^2 - x + 1=0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 1, \ b= -1 \text{ and } c = 1$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{1 + \sqrt{1 - 4 \times 1 \times 1} }{2 \times 1} \text{ and } \beta = \frac{1 - \sqrt{1 - 4 \times 1 \times 1} }{2 \times 1}$

$\displaystyle \Rightarrow \alpha = \frac{1 + \sqrt{-3} }{2} \text{ and } \beta = \frac{1 - \sqrt{-3} }{2}$

$\displaystyle \Rightarrow \alpha = \frac{1 + \sqrt{3}i }{2} \text{ and } \beta = \frac{1 - \sqrt{3}i }{2}$

$\displaystyle \Rightarrow \alpha = \frac{1}{2} + \frac{\sqrt{3}}{2} i \text{ and } \beta = \frac{1}{2} - \frac{\sqrt{3}}{2} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

$\displaystyle \\$

Question 12: $\displaystyle x^2 + x + 1=0$

Answer:

$\displaystyle x^2 + x + 1=0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 1, \ b= 1 \text{ and } c = 1$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{1 - 4 \times 1 \times 1} }{2 \times 1} \text{ and } \beta = \frac{-1 - \sqrt{1 - 4 \times 1 \times 1} }{2 \times 1}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{-3} }{2} \text{ and } \beta = \frac{-1 - \sqrt{-3} }{2}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{3}i }{2} \text{ and } \beta = \frac{-1 - \sqrt{3}i }{2}$

$\displaystyle \Rightarrow \alpha = \frac{-1}{2} + \frac{\sqrt{3}}{2} i \text{ and } \beta = \frac{-1}{2} - \frac{\sqrt{3}}{2} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{-1}{2} \pm \frac{\sqrt{3}}{2} i$

$\displaystyle \\$

Question 13: $\displaystyle 17x^2 -8x + 1 = 0$

Answer:

$\displaystyle 17x^2 -8x + 1 = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 17, \ b= -8 \text{ and } c = 1$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{8 + \sqrt{64 - 4 \times 17 \times 1} }{2 \times 17} \text{ and } \beta = \frac{8 - \sqrt{64 - 4 \times 17 \times 1} }{2 \times 17}$

$\displaystyle \Rightarrow \alpha = \frac{8 + \sqrt{-4} }{34} \text{ and } \beta = \frac{8 - \sqrt{-4} }{34}$

$\displaystyle \Rightarrow \alpha = \frac{8 + 2i }{34} \text{ and } \beta = \frac{8-2i }{34}$

$\displaystyle \Rightarrow \alpha = \frac{4}{17} + \frac{1}{17} i \text{ and } \beta = \frac{4}{17} - \frac{1}{17} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{4}{17} \pm \frac{1}{17} i$

$\displaystyle \\$

Question 14: $\displaystyle 27x^2-10x + 1=0$

Answer:

$\displaystyle 27x^2-10x + 1=0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 27, \ b= -10 \text{ and } c = 1$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{10 + \sqrt{100 - 4 \times 27 \times 1} }{2 \times 27} \text{ and } \beta = \frac{10 - \sqrt{100 - 4 \times 27 \times 1} }{2 \times 27}$

$\displaystyle \Rightarrow \alpha = \frac{10 + \sqrt{-8} }{54} \text{ and } \beta = \frac{10 - \sqrt{-8} }{54}$

$\displaystyle \Rightarrow \alpha = \frac{10 + 2\sqrt{2}i }{54} \text{ and } \beta = \frac{10-2\sqrt{2}i }{54}$

$\displaystyle \Rightarrow \alpha = \frac{5}{27} + \frac{\sqrt{2}}{27} i \text{ and } \beta = \frac{5}{27} - \frac{\sqrt{2}}{17} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{5}{27} \pm \frac{\sqrt{2}}{27} i$

$\displaystyle \\$

Question 15: $\displaystyle 17x^2 + 28x + 12 = 0$

Answer:

$\displaystyle 17x^2 + 28x + 12 = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 17, \ b= 28 \text{ and } c = 12$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{-28 + \sqrt{784 - 4 \times 17 \times 12} }{2 \times 17} \text{ and } \beta = \frac{-28 - \sqrt{784 - 4 \times 17 \times 12} }{2 \times 17}$

$\displaystyle \Rightarrow \alpha = \frac{-28 + \sqrt{-32} }{34} \text{ and } \beta = \frac{-28 - \sqrt{-32} }{34}$

$\displaystyle \Rightarrow \alpha = \frac{-28 + 4\sqrt{2} i }{34} \text{ and } \beta = \frac{-28 - 4\sqrt{2} i }{34}$

$\displaystyle \Rightarrow \alpha = \frac{-14}{17} + \frac{2\sqrt{2}}{17} i \text{ and } \beta = \frac{-14}{17} - \frac{2\sqrt{2}}{17} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{-14}{17} \pm \frac{2\sqrt{2}}{17} i$

$\displaystyle \\$

Question 16: $\displaystyle 21x^2-28x+10=0$

Answer:

$\displaystyle 21x^2-28x+10=0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 21, \ b= -28 \text{ and } c = 10$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{28 + \sqrt{784 - 4 \times 21 \times 10} }{2 \times 21} \text{ and } \beta = \frac{28 - \sqrt{784 - 4 \times 21 \times 10} }{2 \times 21}$

$\displaystyle \Rightarrow \alpha = \frac{28 + \sqrt{-56} }{42} \text{ and } \beta = \frac{28 - \sqrt{-56} }{42}$

$\displaystyle \Rightarrow \alpha = \frac{28 + 2\sqrt{14} i }{42} \text{ and } \beta = \frac{28 - 2\sqrt{14} i }{42}$

$\displaystyle \Rightarrow \alpha = \frac{2}{3} + \frac{\sqrt{14}}{21} i \text{ and } \beta = \frac{2}{3} - \frac{\sqrt{14}}{21} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{2}{3} \pm \frac{\sqrt{14}}{21} i$

$\displaystyle \\$

Question 17: $\displaystyle 8x^2-9x+3=0$

Answer:

$\displaystyle 8x^2-9x+3=0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 8, \ b= -9 \text{ and } c = 3$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{9 + \sqrt{81 - 4 \times 8 \times 3} }{2 \times 8} \text{ and } \beta = \frac{9 - \sqrt{81 - 4 \times 8 \times 3} }{2 \times 8}$

$\displaystyle \Rightarrow \alpha = \frac{9 + \sqrt{-15} }{16} \text{ and } \beta = \frac{9 - \sqrt{-15} }{16}$

$\displaystyle \Rightarrow \alpha = \frac{9 + \sqrt{15} i }{16} \text{ and } \beta = \frac{9 - \sqrt{15} i }{16}$

$\displaystyle \Rightarrow \alpha = \frac{9}{16} + \frac{\sqrt{15}}{16} i \text{ and } \beta = \frac{9}{16} - \frac{\sqrt{15}}{16} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{9}{16} \pm \frac{\sqrt{15}}{16} i$

$\displaystyle \\$

Question 18: $\displaystyle 13x^2+7x+1=0$

Answer:

$\displaystyle 13x^2+7x+1=0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 13, \ b= 7 \text{ and } c = 1$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{-7 + \sqrt{49 - 4 \times 13 \times 1} }{2 \times 13} \text{ and } \beta = \frac{-7 - \sqrt{49 - 4 \times 13 \times 1} }{2 \times 13}$

$\displaystyle \Rightarrow \alpha = \frac{-7 + \sqrt{-3} }{26} \text{ and } \beta = \frac{-7 - \sqrt{-3} }{26}$

$\displaystyle \Rightarrow \alpha = \frac{-7 + \sqrt{3} i }{26} \text{ and } \beta = \frac{-7 - \sqrt{3} i }{26}$

$\displaystyle \Rightarrow \alpha = \frac{-7}{26} + \frac{\sqrt{3}}{26} i \text{ and } \beta = \frac{-7}{26} - \frac{\sqrt{3}}{26} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{-7}{26} \pm \frac{\sqrt{3}}{26} i$

$\displaystyle \\$

Question 19: $\displaystyle 2x^2+x+1=0$

Answer:

$\displaystyle 2x^2+x+1=0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 2, \ b= 1 \text{ and } c = 1$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{1 - 4 \times 2 \times 1} }{2 \times 2} \text{ and } \beta = \frac{-1 - \sqrt{1 - 4 \times 2 \times 1} }{2 \times 2}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{-7} }{4} \text{ and } \beta = \frac{-1 - \sqrt{-7} }{4}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{7} i }{4} \text{ and } \beta = \frac{-1 - \sqrt{7} i }{4}$

$\displaystyle \Rightarrow \alpha = \frac{-1}{4} + \frac{\sqrt{7}}{4} i \text{ and } \beta = \frac{-1}{4} - \frac{\sqrt{7}}{4} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{-1}{4} \pm \frac{\sqrt{7}}{4} i$

$\displaystyle \\$

Question 20: $\displaystyle \sqrt{3} x^2 -\sqrt{2} x + 3 \sqrt{3} = 0$

Answer:

$\displaystyle \sqrt{3} x^2 -\sqrt{2} x + 3 \sqrt{3} = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = \sqrt{3}, \ b= -\sqrt{2} \text{ and } c = 3\sqrt{3}$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{\sqrt{2} + \sqrt{2 - 4 \times \sqrt{3} \times 3\sqrt{3}} }{2 \times \sqrt{3}} \text{ and } \beta = \frac{\sqrt{2} - \sqrt{2 - 4 \times \sqrt{3} \times 3\sqrt{3}} }{2 \times \sqrt{3}}$

$\displaystyle \Rightarrow \alpha = \frac{\sqrt{2} + \sqrt{-34} }{2\sqrt{3}} \text{ and } \beta = \frac{\sqrt{2} - \sqrt{-34} }{2\sqrt{3}}$

$\displaystyle \Rightarrow \alpha = \frac{\sqrt{2} + \sqrt{34} i }{2\sqrt{3}} \text{ and } \beta = \frac{\sqrt{2} - \sqrt{34} i }{2\sqrt{3}}$

$\displaystyle \Rightarrow \alpha = \frac{1}{\sqrt{6}} + \frac{\sqrt{17}}{\sqrt{6}} i \text{ and } \beta = \frac{1}{\sqrt{6}} - \frac{\sqrt{17}}{\sqrt{6}} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{1}{\sqrt{6}} \pm \frac{\sqrt{17}}{\sqrt{6}} i$

$\displaystyle \\$

Question 21: $\displaystyle \sqrt{2} x^2 + x + \sqrt{2} = 0$

Answer:

$\displaystyle \sqrt{2} x^2 + x + \sqrt{2} = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = \sqrt{2}, \ b= 1 \text{ and } c = \sqrt{2}$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{1 - 4 \times \sqrt{2} \times \sqrt{2}} }{2 \times \sqrt{2}} \text{ and } \beta = \frac{-1 - \sqrt{1 - 4 \times \sqrt{2} \times \sqrt{2}} }{2 \times \sqrt{2}}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{-7} }{2\sqrt{2}} \text{ and } \beta = \frac{-1 - \sqrt{-7} }{2\sqrt{2}}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{7} i }{2\sqrt{2}} \text{ and } \beta = \frac{-1 - \sqrt{7} i }{2\sqrt{2}}$

$\displaystyle \Rightarrow \alpha = \frac{-1}{2\sqrt{2}} + \frac{\sqrt{7}}{2\sqrt{2}} i \text{ and } \beta = \frac{-1}{2\sqrt{2}} - \frac{\sqrt{7}}{2\sqrt{2}} i$

$\displaystyle \text{Hence the roots of the equation are } \frac{-1}{2\sqrt{2}} \pm \frac{\sqrt{7}}{2\sqrt{2}} i$

$\displaystyle \\$

$\displaystyle \text{Question 22: } x^2 + x + \frac{1}{\sqrt{2}} = 0$

Answer:

$\displaystyle x^2 + x + \frac{1}{\sqrt{2}} = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 1, \ b= 1 \text{ and } c = \frac{1}{\sqrt{2}}$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{1 - 4 \times 1 \times \frac{1}{\sqrt{2}} }}{2 \times 1} \text{ and } \beta = \frac{-1 - \sqrt{ 1 - 4 \times 1 \times \frac{1}{\sqrt{2} } }}{2 \times 1}$

$\displaystyle \Rightarrow \alpha = \frac{-1 - 2\sqrt{2} }{2} \text{ and } \beta = \frac{-1 - 2\sqrt{2} }{2}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{2\sqrt{2} -1} i }{2} \text{ and } \beta = \frac{-1 - \sqrt{2\sqrt{2} -1} i }{2}$

$\displaystyle \text{Hence the roots of the equation are } \frac{-1 \pm \sqrt{2\sqrt{2} -1} i }{2}$

$\displaystyle \\$

$\displaystyle \text{Question 23: } x^2 + \frac{x}{\sqrt{2}} + 1 = 0$

Answer:

$\displaystyle x^2 + \frac{x}{\sqrt{2}} + 1 = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 1, \ b= \frac{1}{\sqrt{2}} \text{ and } c = 1$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{-\frac{1}{\sqrt{2}} + \sqrt{\frac{1}{2} - 4 \times 1 \times 1} }{2 \times 1} \text{ and } \beta = \frac{-\frac{1}{\sqrt{2}} - \sqrt{\frac{1}{2} - 4 \times 1 \times 1} }{2 \times 1}$

$\displaystyle \Rightarrow \alpha = \frac{-\frac{1}{\sqrt{2}} + \sqrt{\frac{-7}{2}} }{2} \text{ and } \beta = \frac{-\frac{1}{\sqrt{2}} - \sqrt{\frac{-7}{2}} }{2}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + i\sqrt{7} }{2\sqrt{2}} \text{ and } \beta = \frac{-1 - i\sqrt{7} }{2\sqrt{2}}$

$\displaystyle \text{Hence the roots of the equation are } \frac{-1 \pm i\sqrt{7} i }{2\sqrt{2}}$

$\displaystyle \\$

Question 24: $\displaystyle \sqrt{5} x^2 + x + \sqrt{5} = 0$

Answer:

$\displaystyle \sqrt{5} x^2 + x + \sqrt{5} = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = \sqrt{5}, \ b= 1 \text{ and } c = \sqrt{5}$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{1 - 4 \times \sqrt{5} \times \sqrt{5}} }{2 \times \sqrt{5}} \text{ and } \beta = \frac{-1 - \sqrt{1 - 4 \times \sqrt{5} \times \sqrt{5}} }{2 \times \sqrt{5}}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{-19} }{2\sqrt{5}} \text{ and } \beta = \frac{-1 - \sqrt{-19} }{2\sqrt{5}}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{19} i }{2\sqrt{5}} \text{ and } \beta = \frac{-1 - \sqrt{19} i }{2\sqrt{5}}$

$\displaystyle \text{Hence the roots of the equation are } \frac{-1 \pm \sqrt{19} i }{2\sqrt{5}}$

$\displaystyle \\$

Question 25: $\displaystyle -x^2 + x - 2 = 0$

Answer:

$\displaystyle -x^2 + x - 2 = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = -1, \ b= 1 \text{ and } c = -2$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{1 - 4 \times (-1) \times (-2)} }{2 \times (-1)} \text{ and } \beta = \frac{-1 - \sqrt{-1 - 4 \times (-1) \times (-1)} }{2 \times (-1)}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{-7} }{-2} \text{ and } \beta = \frac{-1 - \sqrt{-7} }{-2}$

$\displaystyle \Rightarrow \alpha = \frac{-1 + \sqrt{7} i }{-2} \text{ and } \beta = \frac{-1 - \sqrt{7} i }{-2}$

$\displaystyle \text{Hence the roots of the equation are } \frac{-1 \pm \sqrt{7} i }{-2} \text{ or } \frac{1 \mp \sqrt{7} i }{2}$

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$\displaystyle \text{Question 26: } x^2 - 2 x + \frac{3}{2} = 0$

Answer:

$\displaystyle x^2 - 2 x + \frac{3}{2} = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 1, \ b= -2 \text{ and } c = \frac{3}{2}$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{2 + \sqrt{(-2)^2 - 4 \times 1 \times (\frac{3}{2})} }{2 \times 1} \text{ and } \frac{2 - \sqrt{(-2)^2 - 4 \times 1 \times (\frac{3}{2})} }{2 \times 1}$

$\displaystyle \Rightarrow \alpha = \frac{2 + \sqrt{-2} }{2} \text{ and } \beta = \frac{2 - \sqrt{-2} }{2}$

$\displaystyle \Rightarrow \alpha = \frac{2 + \sqrt{2} i }{2} \text{ and } \beta = \frac{2 - \sqrt{2} i }{2}$

$\displaystyle \Rightarrow \alpha = 1 + \frac{1}{\sqrt{2}} i \text{ and } 1 - \frac{1}{\sqrt{2}} i$

$\displaystyle \text{Hence the roots of the equation are } 1 + \frac{1}{\sqrt{2}} i \text{ and } 1 - \frac{1}{\sqrt{2}} i$

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$\displaystyle \text{Question 27: } 3x^2 - 4x + \frac{20}{3} = 0$

Answer:

$\displaystyle 3x^2 - 4x + \frac{20}{3} = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 3, \ b= -4 \text{ and } c = \frac{20}{3}$

Substituting these values in

$\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \text{ and } \beta = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \Rightarrow \alpha = \frac{4 + \sqrt{16 - 4 \times 3 \times \frac{20}{3} } }{2 \times 3} \text{ and } \beta = \frac{4 - \sqrt{16 - 4 \times 3 \times \frac{20}{3}} }{2 \times 3}$