Exercise 5.1: Quadratic equations of one variable and their solutions.
Formula: For a standard quadratic (2nd-degree) equation $ax^2 + bx + c = 0$, the roots are given by the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
1. Solve: $2x^2 + 9x + 9 = 0$
Solution: Comparing $2x^2 + 9x + 9 = 0$ with the standard quadratic equation $ax^2 + bx + c = 0$, we get $a = 2$, $b = 9$, $c = 9$
Therefore, the two roots are:
$$x = \frac{-9 \pm \sqrt{9^2 – 4(2)(9)}}{2(2)}$$
$$x = \frac{-9 \pm \sqrt{81 – 72}}{4}$$
$$x = \frac{-9 \pm \sqrt{9}}{4}$$
$$x = \frac{-9 \pm 3}{4}$$
Taking the positive sign:
$$x_1 = \frac{-9 + 3}{4} = \frac{-6}{4} = -\frac{3}{2}$$
Taking the negative sign:
$$x_2 = \frac{-9 – 3}{4} = \frac{-12}{4} = -3$$
Required Solution:$x = -\frac{3}{2}, -3$
2. Solve: $3 – 4x – 2x^2 = 0$
Solution: Comparing $3 – 4x – 2x^2 = 0$ with the standard quadratic equation $ax^2 + bx + c = 0$, we get $a = -2$, $b = -4$, $c = 3$
Therefore, the two roots are:
$$x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(-2)(3)}}{2(-2)}$$
$$x = \frac{4 \pm \sqrt{16 + 24}}{-4}$$
$$x = \frac{4 \pm \sqrt{40}}{-4}$$
$$x = \frac{4 \pm 2\sqrt{10}}{-4}$$
$$x = \frac{2(2 \pm \sqrt{10})}{-4}$$
$$x = \frac{-(2 \pm \sqrt{10})}{2} = \frac{-2 \mp \sqrt{10}}{2}$$
Required Solution:$x = \frac{-2 + \sqrt{10}}{2}, \frac{-2 – \sqrt{10}}{2}$
3. Solve: $4x – 1 – x^2 = 0$
Solution: Comparing $4x – 1 – x^2 = 0$ with the standard quadratic equation $ax^2 + bx + c = 0$, we get $a = -1$, $b = 4$, $c = -1$
Therefore, the two roots are:
$$x = \frac{-4 \pm \sqrt{4^2 – 4(-1)(-1)}}{2(-1)}$$
$$x = \frac{-4 \pm \sqrt{16 – 4}}{-2}$$
$$x = \frac{-4 \pm \sqrt{12}}{-2}$$
$$x = \frac{-4 \pm 2\sqrt{3}}{-2}$$
$$x = 2 \mp \sqrt{3}$$
Required Solution:$x = 2 + \sqrt{3}, 2 – \sqrt{3}$
4. Solve: $2x^2 – 5x – 1 = 0$
Solution: Comparing $2x^2 – 5x – 1 = 0$ with the standard quadratic equation $ax^2 + bx + c = 0$, we get $a = 2$, $b = -5$, $c = -1$
Therefore, the two roots are:
$$x = \frac{-(-5) \pm \sqrt{(-5)^2 – 4(2)(-1)}}{2(2)}$$
$$x = \frac{5 \pm \sqrt{25 + 8}}{4}$$
$$x = \frac{5 \pm \sqrt{33}}{4}$$
Required Solution:$x = \frac{5 + \sqrt{33}}{4}, \frac{5 – \sqrt{33}}{4}$
5. Solve: $3x^2 + 7x + 1 = 0$
Solution: Comparing $3x^2 + 7x + 1 = 0$ with the standard quadratic equation $ax^2 + bx + c = 0$, we get $a = 3$, $b = 7$, $c = 1$
Using the formula:
$$x = \frac{-7 \pm \sqrt{(7)^2 – 4(3)(1)}}{2(3)}$$
$$x = \frac{-7 \pm \sqrt{49 – 12}}{6}$$
$$x = \frac{-7 \pm \sqrt{37}}{6}$$
Required Solution:$x = \frac{-7 + \sqrt{37}}{6}, \frac{-7 – \sqrt{37}}{6}$
6. Solve: $2 – 3x^2 + 9x = 0$
Solution: Comparing $2 – 3x^2 + 9x = 0$ with the standard quadratic equation $ax^2 + bx + c = 0$, we get $a = -3$, $b = 9$, $c = 2$
Therefore, the two roots are:
$$x = \frac{-9 \pm \sqrt{9^2 – 4(-3)(2)}}{2(-3)}$$
$$x = \frac{-9 \pm \sqrt{81 + 24}}{-6}$$
$$x = \frac{-9 \pm \sqrt{105}}{-6}$$
$$x = \frac{9 \mp \sqrt{105}}{6}$$
Required Solution:$x = \frac{9 + \sqrt{105}}{6}, \frac{9 – \sqrt{105}}{6}$
7. Solve: $x^2 – 8x + 16 = 0$
Solution: Comparing $x^2 – 8x + 16 = 0$ with the standard quadratic equation $ax^2 + bx + c = 0$, we get $a = 1$, $b = -8$, $c = 16$
Therefore, the two roots are:
$$x = \frac{-(-8) \pm \sqrt{(-8)^2 – 4(1)(16)}}{2(1)}$$
$$x = \frac{8 \pm \sqrt{64 – 64}}{2}$$
$$x = \frac{8 \pm 0}{2} = 4$$
Required Solution:$x = 4$
8. Solve: $2x^2 + 7x – 1 = 0$
Solution: Comparing $2x^2 + 7x – 1 = 0$ with the standard quadratic equation $ax^2 + bx + c = 0$, we get $a = 2$, $b = 7$, $c = -1$
Therefore, the two roots are:
$$x = \frac{-7 \pm \sqrt{7^2 – 4(2)(-1)}}{2(2)}$$
$$x = \frac{-7 \pm \sqrt{49 + 8}}{4}$$
$$x = \frac{-7 \pm \sqrt{57}}{4}$$
Required Solution: $x = \frac{-7 + \sqrt{57}}{4}, \frac{-7 – \sqrt{57}}{4}$
9. Solve: $7x – 2 – 3x^2 = 0$
Solution: Comparing $7x – 2 – 3x^2 = 0$ with the standard quadratic equation $ax^2 + bx + c = 0$, we get $a = -3$, $b = 7$, $c = -2$
Therefore, the two roots are:
$$x = \frac{-7 \pm \sqrt{7^2 – 4(-3)(-2)}}{2(-3)}$$
$$x = \frac{-7 \pm \sqrt{49 – 24}}{-6}$$
$$x = \frac{-7 \pm \sqrt{25}}{-6}$$
$$x = \frac{-7 \pm 5}{-6}$$
Taking the positive sign:
$$x_1 = \frac{-7 + 5}{-6} = \frac{-2}{-6} = \frac{1}{3}$$
Taking the negative sign:
$$x_2 = \frac{-7 – 5}{-6} = \frac{-12}{-6} = 2$$
Required Solution:$x = \frac{1}{3}, 2$