Solve both equations and form a new equation in variable $z$ using the highest root of equation 1 and the lowest root of equation 2 as roots of the new equation.

1. $2 - \frac{11}{x} + \frac{9}{x^2} = 0$
2. $(y - 2)^2 = 2 \frac{1}{4}$

What will be the new equation?

Question

Solve both equations and form a new equation in variable $z$ using the highest root of equation 1 and the lowest root of equation 2 as roots of the new equation.

1. $2 - \frac{11}{x} + \frac{9}{x^2} = 0$
2. $(y - 2)^2 = 2 \frac{1}{4}$

What will be the new equation?

Accepted Answer

**Solution:**

1. Solve the first equation:

Given:
$$2 - \frac{11}{x} + \frac{9}{x^2} = 0$$
Multiply both sides by $x^2$ to clear denominators:

$$2x^2 - 11x + 9 = 0$$

Factorize:

$$2x^2 - 2x - 9x + 9 = 0$$
$$2x(x - 1) - 9(x - 1) = 0$$
$$(2x - 9)(x - 1) = 0$$

Roots are:

$$x = 1, \quad x = \frac{9}{2}$$

The highest root of equation 1 is:

$$\frac{9}{2} = 4.5$$

2. Solve the second equation:

Given:
$$(y - 2)^2 = 2 \frac{1}{4} = \frac{9}{4}$$

Taking square roots:

$$y - 2 = \pm \frac{3}{2}$$

So,

$$y = 2 \pm \frac{3}{2}$$

Roots are:

$$y = 2 + \frac{3}{2} = \frac{7}{2} = 3.5$$
$$y = 2 - \frac{3}{2} = \frac{1}{2} = 0.5$$

The lowest root of equation 2 is:

$$\frac{1}{2} = 0.5$$

3. Form the new quadratic equation with roots $z_1 = \frac{9}{2}$ and $z_2 = \frac{1}{2}$:

Sum of roots $s$:

$$s = \frac{9}{2} + \frac{1}{2} = 5$$

Product of roots $p$:

$$p = \frac{9}{2} 	imes \frac{1}{2} = \frac{9}{4}$$

The quadratic equation with roots $z_1$ and $z_2$ is:

$$z^2 - sz + p = 0$$

Substitute values:

$$z^2 - 5z + \frac{9}{4} = 0$$

Multiply through by 4 to clear denominator:

$$4z^2 - 20z + 9 = 0$$

**Answer:** $4z^2 - 20z + 9 = 0$ (Option b)

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