The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$. Use the common ratio to show that $x^3 - 17x^2 + 80x - 64 = 0$. If $f(x) = x^3 - 17x^2 + 80x - 64$, $x \in \mathbb{R}$, show that $f(1) = 0$, and find another value of $x$ for which $f(x) = 0$. In the case of one of the values of $x$ from part (b), the terms in part (a) will generate a geometric series with a finite sum to infinity. Find this value of $x$ and hence find the sum to infinity.

Leaving Cert Mathematics Algebra: The first three terms of a geome

The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 2 - 2018

Question 2

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The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$. Use the common ratio to show that $x^3 - 17x^2 + 80x - 64 = ... show full transcript

Worked Solution & Example Answer:The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 2 - 2018

Step 1

Show that $x^3 - 17x^2 + 80x - 64 = 0$

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Answer

To find the common ratio $r$ between the terms, we can use:

$r = \frac{5x - 8}{x^2} = \frac{x + 8}{5x - 8}$

Cross-multiplying gives:

$(5x - 8)^2 = x^2(x + 8)$

Expanding both sides leads to:

$25x^2 - 80x + 64 = x^3 + 8x^2$

Bringing all terms to one side leads to:

$x^3 - 17x^2 + 80x - 64 = 0$

Step 2

Show that $f(1) = 0$

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Answer

Substituting $x = 1$ into the function $f(x)$ :

$f(1) = 1^3 - 17(1)^2 + 80(1) - 64$

This simplifies to:

$f(1) = 1 - 17 + 80 - 64 = 0$

Now, to find another value of $x$ for which $f(x) = 0$ , we factor $f(x)$ using synthetic division:

Dividing by $(x - 1)$ gives:

$f(x) = (x - 1)(x^2 - 16)$

The quadratic factor can be further factored as:

$(x - 1)(x - 4)(x + 4)$

Thus, the additional values of $x$ are $x = 4$ and $x = -4$ .

Step 3

Find this value of $x$ and hence find the sum to infinity

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Answer

The finite geometric series occurs when the common ratio $r$ is between -1 and 1. From part (a), if we take $x = 4$ , the terms are:

$x^2 = 16$
$5x - 8 = 12$
$x + 8 = 12$

The common ratio is:

$r = \frac{12}{16} = \frac{3}{4}$

This falls within the acceptable range, thus it will converge. The sum to infinity $S_\infty$ is calculated using the formula:

$S_\infty = \frac{a}{1 - r}$

Where $a$ is the first term ( $16$ ) and $r = \frac{3}{4}$ :

$S_\infty = \frac{16}{1 - \frac{3}{4}} = \frac{16}{\frac{1}{4}} = 64$

The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 2 - 2018

Worked Solution & Example Answer:The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 2 - 2018

Show that $x^3 - 17x^2 + 80x - 64 = 0$

Show that $f(1) = 0$

Find this value of $x$ and hence find the sum to infinity

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