Question

Given that alpha and beta are the roots of the equation 2xsquare - 4x + 3 = 0. Form a new quadratic equation whose root are :
A. 1/alpha, 1/beta
B. 2alpha - 1/beta, 2 beta^-1/2
C. 2alpha, 2beta

Answer 1

Answer to Question #88847 – Math – Algebra

Question

Given that alpha and beta are the roots of the equation $2x$ square $-4x + 3 = 0$. Form a new quadratic equation whose root are:

A. 1/alpha, 1/beta

B. 2alpha - 1/beta, 2 beta^-1/2

C. 2alpha, 2beta

Solution

$$2x^2 - 4x + 3 = 0$$

Let $\alpha, \beta$ be the roots.

$$\text{then}, \alpha + \beta = -\frac{b}{a} = -\frac{-4}{2} = 2$$ $$\alpha\beta = \frac{c}{a} = \frac{3}{2}.$$

A.

Let $\frac{1}{\alpha}, \frac{1}{\beta}$ be the roots of $px^2 + qx + r = 0$ or $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$

then, $\frac{1}{\alpha} + \frac{1}{\beta} = -\frac{q}{p}$ and $\frac{1}{\alpha} \pm \frac{1}{\beta} = \frac{r}{p}$

$$\frac{\alpha + \beta}{\alpha\beta} = -\frac{q}{p} \quad \text{and} \quad \frac{1}{\alpha\beta} = \frac{r}{p}$$ $$\frac{2}{3/2} = -\frac{q}{p} \quad \text{and} \quad \frac{1}{3/2} = \frac{r}{p}$$ $$-\frac{4}{3} = \frac{q}{p} \quad \text{and} \quad \frac{2}{3} = \frac{r}{p}$$

∴ Equation is: $x^2 - \frac{4}{3}x + \frac{2}{3} = 0$

or, $3x^2 - 4x + 2 = 0$.

B.

Let $(2\alpha - \frac{1}{\beta})$ and $(2\beta - \frac{1}{\alpha})$ be the roots of $p x^{2} + q x + r = 0$ or $x^{2} + \frac{q}{p} x + \frac{r}{p} = 0$

then, $2\alpha - \frac{1}{\beta} + 2\beta - \frac{1}{\alpha} = -\frac{q}{p}$ and $(2\alpha - \frac{1}{\beta})(2\beta - \frac{1}{\alpha}) = \frac{r}{p}$

$$2 (\alpha + \beta) - \left(\frac {1}{\beta} + \frac {1}{\alpha}\right) = - \frac {q}{p} \text{ and } 4 \alpha \beta - 2 - 2 + \frac {1}{\alpha \beta} = \frac {r}{p}$$ $$2 (2) - \left(\frac {4}{3}\right) = - \frac {q}{p} \text{ and } 4 \left(\frac {3}{2}\right) - 4 + \frac {1}{3 / 2} = \frac {r}{p}$$ $$4 - \frac {4}{3} = - \frac {q}{p} \text{ and } 6 - 4 + \frac {2}{3} = \frac {r}{p}$$ $$- \frac {8}{3} = \frac {q}{p} \text{ and } \frac {8}{3} = \frac {r}{p}$$

∴ Equation is: $x^{2} - \frac{8}{3} x + \frac{8}{3} = 0$

or, $3x^{2} - 8x + 8 = 0$.

C.

Let $(2\alpha)$ and $(2\beta)$ be the roots of $px^2 + qx + r = 0$ or $x^2 + \frac{q}{p} x + \frac{r}{p} = 0$

then, $2\alpha + 2\beta = -\frac{q}{p}$ and $(2\alpha)(2\beta) = \frac{r}{p}$

$$\begin{array}{l} 2(\alpha + \beta) = -\frac{q}{p} \text{ and } 4\alpha\beta = \frac{r}{p} \\ 2(2) = -\frac{q}{p} \text{ and } 4\left(\frac{3}{2}\right) = \frac{r}{p} \\ 4 = -\frac{q}{p} \text{ and } 6 = \frac{r}{p} \\ -4 = \frac{q}{p} \text{ and } 6 = \frac{r}{p} \\ \end{array}$$

$\therefore$ Equation is: $x^2 - 4x + 6 = 0$

or, $x^2 - 4x + 6 = 0$.

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