Which two values of $x$ are roots of the polynomial below?
$x^2-11 x+13$
A. $x=2.5$
B. $x=\frac{11-\sqrt{-109}}{4}$
C. $x=\frac{11-\sqrt{69}}{2}$
D. $x=\frac{11+\sqrt{-109}}{4}$
E. $x=3$
F. $x=\frac{11+\sqrt{69}}{2}$
Answer (1)
Apply the quadratic formula x = 2 a − b ± b 2 − 4 a c with a = 1 , b = − 11 , and c = 13 .
Substitute the values into the formula: x = 2 11 ± ( − 11 ) 2 − 4 ( 1 ) ( 13 ) .
Simplify the expression: x = 2 11 ± 121 − 52 = 2 11 ± 69 .
The roots are x = 2 11 − 69 and x = 2 11 + 69 , so the final answer is 2 11 − 69 , 2 11 + 69 .
Explanation
Understanding the Problem We are given a quadratic polynomial x 2 − 11 x + 13 and six possible values for its roots. Our objective is to identify which two of the given values are indeed the roots of the polynomial.
Applying the Quadratic Formula We will use the quadratic formula to find the roots of the quadratic polynomial x 2 − 11 x + 13 . The quadratic formula is given by x = 2 a − b ± b 2 − 4 a c where a = 1 , b = − 11 , and c = 13 .
Substitution Substituting the values of a , b , and c into the quadratic formula, we get x = 2 ( 1 ) − ( − 11 ) ± ( − 11 ) 2 − 4 ( 1 ) ( 13 )
Simplification Simplifying the expression: x = 2 11 ± 121 − 52 x = 2 11 ± 69
Identifying the Correct Options Comparing the calculated roots with the given options, we find that the roots are x = 2 11 − 69 and x = 2 11 + 69 These correspond to options C and F.
Final Answer The two values of x that are roots of the polynomial x 2 − 11 x + 13 are 2 11 − 69 and 2 11 + 69 .
Examples
Understanding quadratic equations and their roots is crucial in various fields, such as physics and engineering. For instance, when analyzing the trajectory of a projectile, the roots of a quadratic equation can determine the points where the projectile hits the ground. By knowing the roots, engineers can design systems to accurately predict and control the motion of objects, ensuring safety and efficiency. This principle applies to many real-world scenarios, from designing bridges to optimizing sports equipment.
