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(1) \[x-\dfrac{1}{x}=3,x\ne 0\]

(2) \[\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30},x\ne -4,7\]

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(1) We know the general quadratic equation is given as \[a{{x}^{2}}+bx+c=0\] where a, b are the coefficients of \[{{x}^{2}}\] and x respectively and c is the constant in the equation.

By expressing the given equation \[x-\dfrac{1}{x}=3\] in the form of a general quadratic equation we get,

Multiplying with x on both sides of the equation,

\[{{x}^{2}}-1=3x\]

By subtracting with 3x on both sides we get,

\[{{x}^{2}}-3x-1=0\]

Comparing the above equation with the general quadratic equation we get the values of a, b, c as,

\[a=1\], \[b=-3\], \[c=-1\]

We know the formula to find the roots of a general quadratic equation is given as \[\dfrac{-b\pm \sqrt{D}}{2a}\] where \[D={{b}^{2}}-4ac\] is known as the discriminant.

Applying the above formula by substituting the values of a, b, c to find the discriminant,

For the roots to be real and unequal the discriminant must be greater than 0.

\[\left[ {{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right) \right]>0\]

Squaring \[-3\] and multiplying \[-4\] with\[-1\] we get,

\[\left( 9+4 \right)>0\]

Adding 9 with 4 the resultant is greater than 0,

13 > 0

Hence the roots are real and unequal.

Applying the formula to find the roots of a general quadratic equation by substituting the values of a, b, c and the value of the discriminant \[{{b}^{2}}-4ac\] we get,

\[\dfrac{-\left( -3 \right)\pm \sqrt{13}}{2}\]

By taking square root of 13 and simplifying the expression for \[\dfrac{-b+\sqrt{D}}{2a}\] and \[\dfrac{-b-\sqrt{D}}{2a}\] we get,

\[\begin{align}

& x=3.3027 \\

& x=-0.3027 \\

\end{align}\]

Hence the roots of the equation \[x-\dfrac{1}{x}=3\] are 3.3027 and -0.3027.

(2)

By expressing the given equation \[\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30}\] in the form of a general quadratic equation we get,

Taking a common denominator \[\left( x+4 \right)\left( x-7 \right)\] in the LHS,

\[\dfrac{\left( x-7 \right)-\left( x+4 \right)}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\]

Simplifying the above equation,

\[\dfrac{x-7-x-4}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}\]

\[\dfrac{-11}{{{x}^{2}}-7x+4x-28}=\dfrac{11}{30}\]

Cross multiplying the terms to express as a quadratic equation,

Cancelling out 11 and multiplying with \[-1\] on both sides,

\[{{x}^{2}}-3x-28=-30\]

Expressing the equation in the simplest form possible,

\[{{x}^{2}}-3x+2=0\]

Comparing the above equation with the general quadratic equation we get the values of a, b, c as,

\[a=1\], \[b=-3\], \[c=2\]

We know the formula to find the roots of a general quadratic equation is given as \[\dfrac{-b\pm \sqrt{D}}{2a}\] where \[D={{b}^{2}}-4ac\] is known as the discriminant.

Applying the above formula by substituting the values of a, b, c to find the discriminant,

For the roots to be real and unequal the discriminant must be greater than 0.

\[\left[ {{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( 2 \right) \right]>0\]

Squaring \[-3\] and multiplying \[-4\] with 2 we get,

\[\left( 9-8 \right)>0\]

Subtracting 8 from 9 the resultant is greater than 0,

1 > 0

Hence the roots are real and unequal.

Applying the formula to find the roots of a general quadratic equation by substituting the values of a, b, c and the value of the discriminant \[{{b}^{2}}-4ac\] we get,

\[\dfrac{-\left( -3 \right)\pm \sqrt{1}}{2}\]

By taking square root of 1 and simplifying the expression for \[\dfrac{-b+\sqrt{D}}{2a}\] and \[\dfrac{-b-\sqrt{D}}{2a}\] we get,

x = 2

x = 1

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