Question

1. 3x² - 3 = 0
2. 2x² + 3x - 4 = 0
3. x² + 2 = - 12x
4. 2x² - 22 = 0
5. 5x² - 9x = 2

Explanation:

Step1: Identify quadratic - formula

For a quadratic equation $ax^{2}+bx + c = 0$, the solutions are given by $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

Step2: Solve equation (5) $2x^{2}+3x - 4 = 0$

Here $a = 2$, $b = 3$, $c=-4$. First, calculate the discriminant $\Delta=b^{2}-4ac=3^{2}-4\times2\times(-4)=9 + 32 = 41$. Then $x=\frac{-3\pm\sqrt{41}}{4}$.

Step3: Solve equation (6) $x^{2}+12x + 2 = 0$

Here $a = 1$, $b = 12$, $c = 2$. Calculate the discriminant $\Delta=b^{2}-4ac=12^{2}-4\times1\times2=144 - 8 = 136$. Then $x=\frac{-12\pm\sqrt{136}}{2}=-6\pm\sqrt{34}$.

Step4: Solve equation (7) $2x^{2}-22 = 0$

Rewrite it as $2x^{2}=22$ or $x^{2}=11$, so $x=\pm\sqrt{11}$.

Step5: Solve equation (8) $5x^{2}-9x - 2 = 0$

Here $a = 5$, $b=-9$, $c=-2$. Calculate the discriminant $\Delta=b^{2}-4ac=(-9)^{2}-4\times5\times(-2)=81 + 40 = 121$. Then $x=\frac{9\pm\sqrt{121}}{10}=\frac{9\pm11}{10}$. The solutions are $x = 2$ and $x=-\frac{1}{5}$.

Answer:

(5) $x=\frac{-3\pm\sqrt{41}}{4}$
(6) $x=-6\pm\sqrt{34}$
(7) $x=\pm\sqrt{11}$
(8) $x = 2,x=-\frac{1}{5}$