If the roots of $x^{2} - 7x + 10 = 0$ are $\alpha$ and $\beta$, what is the quadratic equation whose roots are

Question

If the roots of $x^{2} - 7x + 10 = 0$ are $\alpha$ and $\beta$, what is the quadratic equation whose roots are $2\alpha$ and $2\beta$?

Examoogle · Accepted Answer

The correct answer is Option A: $x^{2} - 14x + 40 = 0$. If new roots are $k$ times the original roots, replace $x$ with $\frac{x}{k}$ in the original equation. Here, $k=2$, so substitute $x$ with $\frac{x}{2}$.

Step-by-Step Solution & Concepts:
Shortcut Trick: If new roots are $k$ times the original roots, replace $x$ with $\frac{x}{k}$ in the original equation. Here, $k=2$, so substitute $x$ with $\frac{x}{2}$.

Formula Used: $$x^{2} - (	ext{sum of roots})x + (	ext{product of roots}) = 0$$

Given: A quadratic equation x^{2} - 7x + 10 = 0 with roots alpha and beta.

Steps: Identify sum and product of roots: For the quadratic equation $x^{2} - 7x + 10 = 0$, the sum of roots $\alpha + \beta = 7$ and the product $\alpha \beta = 10$.. Determine roots for the new equation: The new roots are $2\alpha$ and $2\beta$. The sum of these new roots is $2(\alpha + \beta)$ and the product is $(2\alpha)(2\beta) = 4(\alpha \beta)$.. Form the new quadratic equation: Using the standard form $x^{2} - (	ext{Sum})x + (	ext{Product}) = 0$, we substitute the calculated sum $14$ and product $40$.

Answer: A is correct because the new sum of roots is $14$ and the new product of roots is $40$, resulting in the quadratic equation $x^{2} - 14x + 40 = 0$.

Exam Tip: This transformation technique is widely used in Theory of Equations and can be extended to higher-degree polynomials by substituting $x$ with $\frac{x}{k}$.

Examoogle · Answer

Option B: $x^{2} - 7x + 20 = 0$

Examoogle · Answer

Option C: $x^{2} - 14x + 20 = 0$

Examoogle · Answer

Option D: $x^{2} - 7x + 40 = 0$

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