Question

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

Answer 1

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 =
`(-b\pm√(b^2-4ac))/(2a)`

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

a = 2, b = -3 and c = -6

x =
`(3\pm√((-3)^2-4(2)(-6)))/(2(2))`

x =
`(3\pm√(9+48))/(4)`

x =
`(3\pm√(57))/(4)`

x =
`(3+√(57))/(4),(3-√(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 =
`(-b\pm√(b^2-4ac))/(2a)`

x =
`(-3\pm√(3^2-4(2)(2)))/(2(2))`

x =
`(-3\pm√(9-16))/(4)`

x =
`(-3\pm i√(7))/(4)`

x =
`(-3+ i√(7))/(4),(-3- i√(7))/(4)`

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