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j137b 2. given that (x^2 - 2x - 5 = 0) has 2 different roots, and (x^2 …

Question

j137b

  1. given that (x^2 - 2x - 5 = 0) has 2 different roots, and (x^2 + ax + b = 0) has 2 roots, each 2 less than the corresponding root of the equation, find the values of (a) and (b).\

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hint\
if the roots of (x^2 - 2x - 5 = 0) are (\alpha) and (\beta), the roots of the second quadratic equation will be (\alpha - 2) and (\beta - 2).

Explanation:

Step1: Identify original roots relation

Let \(\alpha\) and \(\beta\) be roots of \(x^2 - 2x - 5 = 0\).

$$\alpha + \beta = 2, \quad \alpha\beta = -5$$

Step2: Relate to new roots

The roots of \(x^2 + ax + b = 0\) are \(\alpha - 2\) and \(\beta - 2\).

$$(\alpha - 2) + (\beta - 2) = -a, \quad (\alpha - 2)(\beta - 2) = b$$

Step3: Solve for \(a\)

Substitute the sum of roots into the equation.

$$-a = \alpha + \beta - 4 = 2 - 4 = -2 \implies a = 2$$

Step4: Solve for \(b\)

Expand and substitute values to find \(b\).

$$b = \alpha\beta - 2(\alpha + \beta) + 4 = -5 - 2(2) + 4 = -5$$

Answer:

\(a = 2\), \(b = -5\)