# Show that $\sin {36^0}$ is a root of the quadratic equation $16{x^4} - 20{x^2} + 5 = 0$.

Last updated date: 18th Mar 2023

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Answer

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Hint – For solving such a question, use a simple formula of roots of quadratic equation.

Given equation:

$16{x^4} - 20{x^2} + 5 = 0$

Since the power of $x$ is$4\& 2$

So, let${x^2} = t$ in the above equation.

Then the equation becomes:

$16{t^2} - 20t + 5 = 0$

As we know the formula for roots of quadratic equation is:

$\left[ {x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right]$ for any general quadratic equation of the form $a{x^2} + bx + c = 0$

Hence roots of the given quadratic equation are:

$

t = \dfrac{{ - \left( { - 20} \right) \pm \sqrt {{{\left( { - 20} \right)}^2} - \left( {4 \times 16 \times 5} \right)} }}{{2 \times 16}} \\

t = \dfrac{{20 \pm \sqrt {400 - 320} }}{{32}} \\

t = \dfrac{{20 \pm \sqrt {80} }}{{32}} \\

t = \dfrac{{4\left( {5 \pm \sqrt 5 } \right)}}{{32}} \\

t = \dfrac{{\left( {5 \pm \sqrt 5 } \right)}}{8} \\

$

Substituting the value of $x$ in place of $t$ we get:

$

{x^2} = \dfrac{{\left( {5 \pm \sqrt 5 } \right)}}{8} \\

x = \sqrt {\dfrac{{\left( {5 \pm \sqrt 5 } \right)}}{8}} \\

$

Multiplying and dividing numbers inside the root by $2$.

$

x = \sqrt {\dfrac{{2\left( {5 \pm \sqrt 5 } \right)}}{{16}}} \\

x = \dfrac{{\sqrt {10 \pm 2\sqrt 5 } }}{4} \\

$

As we know that

$\sin {36^0} = \dfrac{{\sqrt {10 - 2\sqrt 5 } }}{4}$

Hence, $\sin {36^0}$ is a root of a given quadratic equation.

Note- Whenever you find such type of problems, you can convert your \[4th\] order equation into quadratic equation by assuming some variable as done in the case above, after that with the help of quadratic formula easily evaluate the unknown variable. Formulas of roots of the quadratic equation mentioned above must be remembered in order to solve the quadratic equation easily.

Given equation:

$16{x^4} - 20{x^2} + 5 = 0$

Since the power of $x$ is$4\& 2$

So, let${x^2} = t$ in the above equation.

Then the equation becomes:

$16{t^2} - 20t + 5 = 0$

As we know the formula for roots of quadratic equation is:

$\left[ {x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right]$ for any general quadratic equation of the form $a{x^2} + bx + c = 0$

Hence roots of the given quadratic equation are:

$

t = \dfrac{{ - \left( { - 20} \right) \pm \sqrt {{{\left( { - 20} \right)}^2} - \left( {4 \times 16 \times 5} \right)} }}{{2 \times 16}} \\

t = \dfrac{{20 \pm \sqrt {400 - 320} }}{{32}} \\

t = \dfrac{{20 \pm \sqrt {80} }}{{32}} \\

t = \dfrac{{4\left( {5 \pm \sqrt 5 } \right)}}{{32}} \\

t = \dfrac{{\left( {5 \pm \sqrt 5 } \right)}}{8} \\

$

Substituting the value of $x$ in place of $t$ we get:

$

{x^2} = \dfrac{{\left( {5 \pm \sqrt 5 } \right)}}{8} \\

x = \sqrt {\dfrac{{\left( {5 \pm \sqrt 5 } \right)}}{8}} \\

$

Multiplying and dividing numbers inside the root by $2$.

$

x = \sqrt {\dfrac{{2\left( {5 \pm \sqrt 5 } \right)}}{{16}}} \\

x = \dfrac{{\sqrt {10 \pm 2\sqrt 5 } }}{4} \\

$

As we know that

$\sin {36^0} = \dfrac{{\sqrt {10 - 2\sqrt 5 } }}{4}$

Hence, $\sin {36^0}$ is a root of a given quadratic equation.

Note- Whenever you find such type of problems, you can convert your \[4th\] order equation into quadratic equation by assuming some variable as done in the case above, after that with the help of quadratic formula easily evaluate the unknown variable. Formulas of roots of the quadratic equation mentioned above must be remembered in order to solve the quadratic equation easily.

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