Question

How do you simplify \[\dfrac{{10{k^2} + 55k + 75}}{{20{k^2} - 10k - 150}}\] and find any non permissible values?

Answer 1

Hint: Here in this question, we have to find the simplest form of the given fraction. In give fraction the numerator and denominator having a quadratic equation, first we have to find the factors of the both quadratic equation by using the factorisation and by substituting and further simplification we get the required solution.

Complete step by step answer:
Consider the given fraction:
\[ \Rightarrow \,\,\dfrac{{10{k^2} + 55k + 75}}{{20{k^2} - 10k - 150}}\]-------(1)
Take 5 as common in numerator and 10 as common in denominator.
\[ \Rightarrow \,\,\dfrac{{5\left( {2{k^2} + 11k + 15} \right)}}{{10\left( {2{k^2} - k - 15} \right)}}\]
On simplification, we get
\[ \Rightarrow \,\,\dfrac{{\left( {2{k^2} + 11k + 15} \right)}}{{2\left( {2{k^2} - k - 15} \right)}}\]------(2)
Now, we have to find the factors of quadratic equations \[2{k^2} + 11k + 15\] and \[2{k^2} - k - 15\] by using the method of factorization.
Consider,
\[ \Rightarrow \,\,\,2{k^2} + 11k + 15\]
 Break the middle term as the summation of two numbers such that its product is equal to 30. Calculated above such two numbers are 6 and 5.
\[ \Rightarrow \,\,\,2{k^2} + 6k + 5k + 15\]
Making pairs of terms in the above expression
\[ \Rightarrow \,\,\,\left( {2{k^2} + 6k} \right) + \left( {5k + 15} \right)\]
Take out greatest common divisor GCD from the both pairs, then
\[ \Rightarrow \,\,\,2k\left( {k + 3} \right) + 5\left( {k + 3} \right)\]
Take \[\left( {k + 3} \right)\] common
\[ \Rightarrow \,\,\,\left( {k + 3} \right)\left( {2k + 5} \right)\]-------(3)
Consider,
\[ \Rightarrow \,\,\,2{k^2} - k - 15\]
 Break the middle term as the summation of two numbers such that its product is equal to -30. Calculated above such two numbers are -6 and 5.
\[ \Rightarrow \,\,\,2{k^2} - 6k + 5k - 15\]
Making pairs of terms in the above expression
\[ \Rightarrow \,\,\,\left( {2{k^2} - 6k} \right) + \left( {5k - 15} \right)\]
Take out greatest common divisor GCD from the both pairs, then
\[ \Rightarrow \,\,\,2k\left( {k - 3} \right) + 5\left( {k - 3} \right)\]
Take \[\left( {k - 3} \right)\] common
\[ \Rightarrow \,\,\,\left( {k - 3} \right)\left( {2k + 5} \right)\]-------(4)
Substitute equation (4) and (3) in equation (2), then
\[ \Rightarrow \,\,\dfrac{{\left( {k + 3} \right)\left( {2k + 5} \right)}}{{2\left( {k - 3} \right)\left( {2k + 5} \right)}}\]
Cancelling like terms in both numerator and denominator
\[ \Rightarrow \,\,\dfrac{{\left( {k + 3} \right)}}{{2\left( {k - 3} \right)}}\]; \[k \ne 3\]
This is the required solution.

Note: The equation is a fraction of quadratic equation. This problem can be solved by using the sum product rule. This defines as for the general quadratic equation \[a{x^2} + bx + c\], the product of \[a{x^2}\] and c is equal to the sum of bx of the equation. Hence we obtain the factors. The factors for the equation depend on the degree of the equation, and later we can divide the factors in fraction.