Question

j 137a kumon
root-coefficient relationships

1. given that (x^2 - 3x + 1 = 0) has 2 different roots, create a quadratic equation with roots 1 greater than the corresponding roots of the original equation.

hint
if the roots of (x^2 - 3x + 1 = 0) are (alpha) and (\beta), the roots of the new equation will be (alpha + 1) and (\beta + 1).

Explanation:

Step1: Identify original roots

Let \(\alpha\) and \(\beta\) be the roots of \(x^2 - 3x + 1 = 0\).

Step2: Find sum and product

Using Vieta's formulas:

$$\alpha + \beta = 3$$

$$\alpha\beta = 1$$

Step3: Define new roots

The new roots are \(\alpha + 1\) and \(\beta + 1\).

Step4: Calculate new sum

$$(\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = 3 + 2 = 5$$

Step5: Calculate new product

$$(\alpha + 1)(\beta + 1) = \alpha\beta + (\alpha + \beta) + 1 = 1 + 3 + 1 = 5$$

Step6: Form new equation

$$x^2 - 5x + 5 = 0$$

Answer:

\(x^2 - 5x + 5 = 0\)