Answer:
Step-by-step explanation:
1). Given equation is,
2x² - 3x = 6
2x² - 3x - 6 = 0
To find the solutions of the equation we will use quadratic formula,
x = [tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Substitute the values of a, b and c in the formula,
a = 2, b = -3 and c = -6
x = [tex]\frac{3\pm\sqrt{(-3)^2-4(2)(-6)}}{2(2)}[/tex]
x = [tex]\frac{3\pm\sqrt{9+48}}{4}[/tex]
x = [tex]\frac{3\pm\sqrt{57}}{4}[/tex]
x = [tex]\frac{3+\sqrt{57}}{4},\frac{3-\sqrt{57}}{4}[/tex]
Therefore, there are two real solutions.
2). Given equation is,
x² + 1 = 2x
x² - 2x + 1 = 0
(x - 1)² = 0
x = 1
Therefore, there is one real solution of the equation.
3). 2x² + 3x + 2 = 0
By applying quadratic formula,
x = [tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
x = [tex]\frac{-3\pm\sqrt{3^2-4(2)(2)}}{2(2)}[/tex]
x = [tex]\frac{-3\pm\sqrt{9-16}}{4}[/tex]
x = [tex]\frac{-3\pm i\sqrt{7}}{4}[/tex]
x = [tex]\frac{-3+ i\sqrt{7}}{4},\frac{-3- i\sqrt{7}}{4}[/tex]
Therefore, there are two complex (non real) solutions.