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This is one of our many quizzes on Excluded values. Click on this to access more such competitive quizzes on Algebra.

Excluded values Quiz 2

mathematics

 Which of the subsequent are answers to the equation $\dfrac{{{x^2} - 4}}{{{x^2} + 5x + 6}} = 0$?

 To find the answer to the equation, simply find the factors of the numerator and equate it to zero. Also, remove the values of $x$ at which denominator is zero.

We are given, $\dfrac{{{x^2} - 4}}{{{x^2} + 5x + 6}} = 0$ $ \\ $ Factorize the equation and equate numerator to zero. $ \\ \dfrac{{(x - 2)(x + 2)}}{{(x + 3)(x + 2)}} = 0 \\ $ On equating to zero, $ \\   \Rightarrow x - 2 = 0 \\   \Rightarrow x + 2 = 0 \\  $  So the values of $x$ would be, $ \\ \Rightarrow x = 2 \\   \Rightarrow x =  - 2 \\  $  Also, the values are excluded at which denominator is zero, The denominator is zero when, $ \\  \Rightarrow x =  - 3 \\   \Rightarrow x =  - 2 \\ $ So, $x$ cannot be equal to $ - 2$  Hence, The answer to the equation is $2$ 

 If the common $($arithmetic mean$)$ of $x,y$, and $9$ is $12$, what's the common of $x + 2, y - 6$, and $10$?

 To solve the following question, apply the formula of arithmetic number $\text{Arithmetic mean }= \dfrac{\text{Sum of numbers of a group}}{\text{Number of items in the group}}$ .

We are given, $ \dfrac{{x + y + 9}}{3} = 12 \\ \Rightarrow x + y + 9 = 36 \\ \Rightarrow x + y = 27 \\ $ We need to calculate, $ \\ \dfrac{{x + 2 + y - 6 + 10}}{3} = ? \\ $ Simplifying the equation, $ \Rightarrow \dfrac{{x + y + 6}}{3} = ?$ We know that $x + y = 27$, $ \\ \Rightarrow \dfrac{{27 + 6}}{3} = ? \\   \Rightarrow \dfrac{{33}}{3} = 11 \\  $

 Find out the excluded values of the subsequent algebraically fraction $\dfrac{{2{x^2} - 5x - 12}}{{{x^2} - 12x + 23}}$

 To find the excluded values you need to factor the denominator and find its roots.

We are given, $\dfrac{{2{x^2} - 5x - 12}}{{{x^2} - 12x + 32}}$ In order to find the excluded value of the algebraic fraction, find the value at which denominator is zero. $   \Rightarrow {x^2} - (8 + 4)x + 32 \\   \Rightarrow {x^2} - 8x - 4x + 32 \\   \Rightarrow x(x - 8) - 4(x - 8) \\   \Rightarrow (x - 8)(x - 4) \\  $. The value of denominator is zero when $x = 4,8$  $ \\ $ Hence the excluded values are $4  and  8$ 

 How do you simplify and state the excluded values for $\dfrac{{x + 4}}{{{x^2} - 16}}$

 To find the excluded values, find the values of $x$ at which denominator is equal to zero.

Note that $\dfrac{{x + 4}}{{{x^2} - 16}}$ is simply outlined if the divisor isn't adequate to zero. $ \\ $ That is ${x^2} - 16 \ne 0 \to x = 4$ and $x =  - 4$ are excluded values $ \\ $ If $x \ne  \pm 4$ $ \\ $ Then $\dfrac{{x + 4}}{{{x^2} - 16}}$ $ \\ $ Factor ${x^2} - 16$ $ \\ = \dfrac{{x + 4}}{{(x + 4)(x - 4)}} \\ = \dfrac{1}{{x - 4}} \\  $

 Simplify $\dfrac{{21{x^2} - 39x - 6}}{{9{x^2} + 45x + 54}}$ and state the excluded value.

 In order to change a rational expression, we tend to resolve the dividend and divisor each. Then eliminate the factors that are common in each dividend and divisor. So we have a tendency to get a simplified rational expression.

$  {\text{Factor }}21{x^2} - 39x - 6 \\   = \dfrac{{3\left( {7{x^2} - 13x - 2} \right)}}{{9{x^2} + 45x + 54}} \\  {\text{Factor }}7{x^2} - 13x - 2 \\   = \dfrac{{3\left( {7x + 1} \right)\left( {x - 2} \right)}}{{9{x^2} + 45x + 54}} \\  {\text{Factor }}9{x^2} + 45x + 54 \\   = \dfrac{{3\left( {7x + 1} \right)\left( {x - 2} \right)}}{{9\left( {{x^2} + 5x + 6} \right)}} \\  {\text{Factor }}{x^2} + 5x + 6 \\   = \dfrac{{3\left( {7x + 1} \right)\left( {x - 2} \right)}}{{9\left( {x + 2} \right)\left( {x + 3} \right)}} \\  {\text{Cancel the numbers: }}\dfrac{3}{9} = \dfrac{1}{3} \\   = \dfrac{{\left( {7x + 1} \right)\left( {x - 2} \right)}}{{3\left( {x + 2} \right)\left( {x + 3} \right)}} \\  (x + 2)(x + 3) \\  {\text{Set every part adequate to zero}}{\text{.}} \\  (x + 2) = 0 \\  (x + 3) = 0 \\  x =  - 2 \\  x =  - 3 \\  $