# Solve the following quadratic equation: \[\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30},x\ne 4,7\]

Answer

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int: Simplify the given equation by taking LCM of the terms given in the denominator. Solve the equation on the left hand side and equate it to the right hand side to get the values of variable x which satisfy the given equation.

Complete step-by-step answer:

We have the equation \[\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30},x\ne 4,7\]. We have to find solutions to this equation.

We will simplify the terms on the left hand side of the equation and equate it to the right hand side.

Taking LCM of terms on left side, we have \[\dfrac{\left( x-7 \right)-\left( x+4 \right)}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\].

Further simplifying the equation, we have \[\dfrac{-7-4}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\].

Thus, we have \[\dfrac{-11}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\].

\[\Rightarrow \dfrac{-1}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{1}{30}\]

Cross multiplying the terms on both sides of the equation, we have \[\left( x+4 \right)\left( x-7 \right)=-30\].

Simplifying the above equation, we have \[{{x}^{2}}+4x-7x-28=-30\].

Thus, we have \[{{x}^{2}}-3x+2=0\]. We will now find roots of this quadratic equation using the factorization method by splitting the terms.

We can rewrite the above equation as \[{{x}^{2}}-x-2x+2=0\].

Taking out the common terms, we have \[x\left( x-1 \right)-2\left( x-1 \right)=0\].

So, we have \[\left( x-1 \right)\left( x-2 \right)=0\].

Thus, we have \[x=1,2\].

Hence, the roots of the equation \[\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30},x\ne 4,7\] are \[x=1,2\].

Note: We observe that the equation given to us is a polynomial equation. Polynomial is an expression consisting of variables and coefficients that involves only the operations of addition, subtraction, multiplication or division, and non-negative integer exponents of variables. Degree of a polynomial is the value of the highest power of degrees of its individual term. We observe that the polynomial given to us is of degree 2. There are multiple ways to solve a polynomial equation, like completing the square and factorizing the polynomial by splitting the intermediate terms. We have solved this question using the factorization method by splitting the intermediate terms. It’s necessary to keep in mind that we can’t consider the value of x to be -4 or 7 as the function is not defined at these values. We can check if the calculated solutions are correct or not by substituting the value of solutions in the equation and checking if they satisfy the given equation or not.

Complete step-by-step answer:

We have the equation \[\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30},x\ne 4,7\]. We have to find solutions to this equation.

We will simplify the terms on the left hand side of the equation and equate it to the right hand side.

Taking LCM of terms on left side, we have \[\dfrac{\left( x-7 \right)-\left( x+4 \right)}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\].

Further simplifying the equation, we have \[\dfrac{-7-4}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\].

Thus, we have \[\dfrac{-11}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\].

\[\Rightarrow \dfrac{-1}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{1}{30}\]

Cross multiplying the terms on both sides of the equation, we have \[\left( x+4 \right)\left( x-7 \right)=-30\].

Simplifying the above equation, we have \[{{x}^{2}}+4x-7x-28=-30\].

Thus, we have \[{{x}^{2}}-3x+2=0\]. We will now find roots of this quadratic equation using the factorization method by splitting the terms.

We can rewrite the above equation as \[{{x}^{2}}-x-2x+2=0\].

Taking out the common terms, we have \[x\left( x-1 \right)-2\left( x-1 \right)=0\].

So, we have \[\left( x-1 \right)\left( x-2 \right)=0\].

Thus, we have \[x=1,2\].

Hence, the roots of the equation \[\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30},x\ne 4,7\] are \[x=1,2\].

Note: We observe that the equation given to us is a polynomial equation. Polynomial is an expression consisting of variables and coefficients that involves only the operations of addition, subtraction, multiplication or division, and non-negative integer exponents of variables. Degree of a polynomial is the value of the highest power of degrees of its individual term. We observe that the polynomial given to us is of degree 2. There are multiple ways to solve a polynomial equation, like completing the square and factorizing the polynomial by splitting the intermediate terms. We have solved this question using the factorization method by splitting the intermediate terms. It’s necessary to keep in mind that we can’t consider the value of x to be -4 or 7 as the function is not defined at these values. We can check if the calculated solutions are correct or not by substituting the value of solutions in the equation and checking if they satisfy the given equation or not.

Last updated date: 02nd Oct 2023

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Total views: 362.4k

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