Find the root of $16x - \dfrac{10}{x} = 27$.
Answer
525.3k+ views
Hint: The given equation is a quadratic equation which in not in the standard form so, first we will convert this equation into standard form i.e., $a{x^2} + bx + c = 0$ and then find the root by using the formula method. The formula method to find the root of a standard equation is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Complete step by step solution:
The given equation is $16x - \dfrac{{10}}{x} = 27$
To convert it into standard form,
We multiply both sides by $x$,
$x(16x - \dfrac{{10}}{x}) = 27x$
Solving the bracket,
$16x \times x - x \times \dfrac{{10}}{x} = 27x$
Multiplying the terms,
$16{x^2} - 10 = 27x$
Taking $27x$ to the left side,
$16{x^2} - 27x - 10 = 0$
$\therefore $ The equation in standard form is $16{x^2} - 27x - 10 = 0$.
By comparing it with standard quadratic equation $a{x^2} + bx + c = 0$, we get, $a = 16,b = - 27,c = - 10$.
The roots of the quadratic equation are given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting all the values in this equation,
$x = \dfrac{{ - ( - 27) \pm \sqrt {{{( - 27)}^2} - 4(16)( - 10)} }}{{2(16)}}$
First solving the square root part,
Taking square of $ - 27$,
$x = \dfrac{{ - ( - 27) \pm \sqrt {729 + 4(16)(10)} }}{{2(16)}}$
Multiplying the terms inside the square root,
$x = \dfrac{{ - ( - 27) \pm \sqrt {729 + 640} }}{{2(16)}}$
Adding the terms inside the square root,
$x = \dfrac{{ - ( - 27) \pm \sqrt {1369} }}{{2(16)}}$
Taking square root of $1369$,
$x = \dfrac{{ - ( - 27) \pm 37}}{{2(16)}}$
Multiplying $2$ by \[16\],
$x = \dfrac{{ - ( - 27) \pm 37}}{{32}}$
Multiplying the signs,
$x = \dfrac{{27 \pm 37}}{{32}}$
We can write this as,
$x = \dfrac{{27 + 37}}{{32}}$ and $x = \dfrac{{27 - 37}}{{32}}$
Solving the numerator part,
$x = \dfrac{{64}}{{32}}$ and $x = \dfrac{{ - 10}}{{32}}$
Dividing the numbers,
$x = 2$ and $x = \dfrac{{ - 10}}{{32}}$
Therefore, the roots of $16x - 10/x = 27$ is $x = 2$ and $x = \dfrac{{ - 10}}{{32}}$.
Note: Here, we can even use the factorization to find the roots of a quadratic equation. It is necessary to first check whether the given equation is in the standard form or not, if not then we have to convert it into standard form and solve the equation by any one of the three methods : factorization, formula method or completing the square method.
Complete step by step solution:
The given equation is $16x - \dfrac{{10}}{x} = 27$
To convert it into standard form,
We multiply both sides by $x$,
$x(16x - \dfrac{{10}}{x}) = 27x$
Solving the bracket,
$16x \times x - x \times \dfrac{{10}}{x} = 27x$
Multiplying the terms,
$16{x^2} - 10 = 27x$
Taking $27x$ to the left side,
$16{x^2} - 27x - 10 = 0$
$\therefore $ The equation in standard form is $16{x^2} - 27x - 10 = 0$.
By comparing it with standard quadratic equation $a{x^2} + bx + c = 0$, we get, $a = 16,b = - 27,c = - 10$.
The roots of the quadratic equation are given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting all the values in this equation,
$x = \dfrac{{ - ( - 27) \pm \sqrt {{{( - 27)}^2} - 4(16)( - 10)} }}{{2(16)}}$
First solving the square root part,
Taking square of $ - 27$,
$x = \dfrac{{ - ( - 27) \pm \sqrt {729 + 4(16)(10)} }}{{2(16)}}$
Multiplying the terms inside the square root,
$x = \dfrac{{ - ( - 27) \pm \sqrt {729 + 640} }}{{2(16)}}$
Adding the terms inside the square root,
$x = \dfrac{{ - ( - 27) \pm \sqrt {1369} }}{{2(16)}}$
Taking square root of $1369$,
$x = \dfrac{{ - ( - 27) \pm 37}}{{2(16)}}$
Multiplying $2$ by \[16\],
$x = \dfrac{{ - ( - 27) \pm 37}}{{32}}$
Multiplying the signs,
$x = \dfrac{{27 \pm 37}}{{32}}$
We can write this as,
$x = \dfrac{{27 + 37}}{{32}}$ and $x = \dfrac{{27 - 37}}{{32}}$
Solving the numerator part,
$x = \dfrac{{64}}{{32}}$ and $x = \dfrac{{ - 10}}{{32}}$
Dividing the numbers,
$x = 2$ and $x = \dfrac{{ - 10}}{{32}}$
Therefore, the roots of $16x - 10/x = 27$ is $x = 2$ and $x = \dfrac{{ - 10}}{{32}}$.
Note: Here, we can even use the factorization to find the roots of a quadratic equation. It is necessary to first check whether the given equation is in the standard form or not, if not then we have to convert it into standard form and solve the equation by any one of the three methods : factorization, formula method or completing the square method.
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