For what value of the constant k\(\) does the quadratic \( 2x^2 - 5x + k\) have a double root?
+2446 Rahuan, I am not completely sure what it means either, but I will interpret "double root" as a zero with a multiplicity of two. Assuming the above is true, we can use the property of the discriminant of a quadratic, \(b^2-4ac\) . The discriminant, if equal to zero, results in a quadratic with a "double root."
| \(b^2-4ac=0\) | Plug in the given values of a, b, and c of the quadratic. |
| \((-5)^2-4(2)(k)=0\) | Now, solve for k. |
| \(25-8k=0\) | Subtract 25 from both sides. |
| \(-8k=-25\) | |
| \(k=\frac{25}{8} \) | |
+2446 Rahuan, I am not completely sure what it means either, but I will interpret "double root" as a zero with a multiplicity of two. Assuming the above is true, we can use the property of the discriminant of a quadratic, \(b^2-4ac\) . The discriminant, if equal to zero, results in a quadratic with a "double root."
| \(b^2-4ac=0\) | Plug in the given values of a, b, and c of the quadratic. |
| \((-5)^2-4(2)(k)=0\) | Now, solve for k. |
| \(25-8k=0\) | Subtract 25 from both sides. |
| \(-8k=-25\) | |
| \(k=\frac{25}{8} \) | |
+502 does he mean like finding the value of k and then putting the answer as \(\frac{5^2}{8}\)
+2446 I do not believe so; there is no reason not to simplify the fraction completely.
