Question

Match each equation on the left with the number and type of its solutions on the right.

Answer 1

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 = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

   Substitute the values of a, b and c in the formula,

   a = 2, b = -3 and c = -6

   x = $\frac{3\pm\sqrt{(-3)^2-4(2)(-6)}}{2(2)}$

   x = $\frac{3\pm\sqrt{9+48}}{4}$

   x = $\frac{3\pm\sqrt{57}}{4}$

   x = $\frac{3+\sqrt{57}}{4},\frac{3-\sqrt{57}}{4}$

   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 = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

      x = $\frac{-3\pm\sqrt{3^2-4(2)(2)}}{2(2)}$

      x = $\frac{-3\pm\sqrt{9-16}}{4}$

      x = $\frac{-3\pm i\sqrt{7}}{4}$

      x = $\frac{-3+ i\sqrt{7}}{4},\frac{-3- i\sqrt{7}}{4}$

      Therefore, there are two complex (non real) solutions.