Final answer:
Lindsey's quadratic equation x²−2x+6=0 has complex roots because the discriminant in the quadratic formula is negative, indicating that there are no real-number solutions.
Step-by-step explanation:
Lindsey is solving the quadratic equation x²−2x+6=0. A quadratic equation is generally expressed in the form ax²+bx+c=0, where a, b, and c are coefficients and x represents the variable. To find the roots of the quadratic equation, we can use the quadratic formula, which is stated as:
x = −b ± √(b² −4ac) / (2a)
However, before jumping into the formula, one should always check if the quadratic can be factored easily or if it's a perfect square. In Lindsey's case, the equation does not factor nicely, and it is not a perfect square. So let's apply the quadratic formula:
In the equation x²−2x+6=0, a = 1, b = −2, and c = 6. Plugging these values into the quadratic formula, we get:
x = (2 ± √((-2)² − 4∙(1)∙(6))) / (2∙1)
x = (2 ± √(4 − 24)) / 2
x = (2 ± √(−20)) / 2
This shows that the equation has complex roots because the discriminant (b² − 4ac) is negative. Therefore, the statement that Lindsey's quadratic equation x²−2x+6=0 has complex roots is true.