Show that $\sin {36^0}$ is a root of the quadratic equation $16{x^4} - 20{x^2} + 5 = 0$.
Answer
366.9k+ views
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.
Last updated date: 02nd Oct 2023
•
Total views: 366.9k
•
Views today: 8.66k
Recently Updated Pages
What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE
