Question

The quadratic equation whose roots are \[ \cos72^\circ \quad \text{and} \quad \sin54^\circ \] is

• A. $4x^2+\sqrt{5}x-1=0$
• B. $4x^2+2\sqrt{5}x+1=0$
• C. $4x^2-2\sqrt{5}x+1=0$
• D. $x^2-2\sqrt{5}x+4=0$

Hint:

Whenever trigonometric values are roots of an equation, first convert them into exact algebraic forms before forming the polynomial.

Answer 1

The Correct Option is C

Step 1: Recall the special values.
A quadratic with roots $\alpha,\beta$ is $x^2-(\alpha+\beta)x+\alpha\beta=0$. Here $\cos72^\circ=\dfrac{\sqrt5-1}{4}$ and $\sin54^\circ=\cos36^\circ=\dfrac{\sqrt5+1}{4}$.
Step 2: Name the roots.
Let $\alpha=\dfrac{\sqrt5-1}{4}$ and $\beta=\dfrac{\sqrt5+1}{4}$.
Step 3: Add the roots.
$\alpha+\beta=\dfrac{(\sqrt5-1)+(\sqrt5+1)}{4}=\dfrac{2\sqrt5}{4}=\dfrac{\sqrt5}{2}$.
Step 4: Multiply the roots.
$\alpha\beta=\dfrac{(\sqrt5-1)(\sqrt5+1)}{16}=\dfrac{5-1}{16}=\dfrac14$.
Step 5: Write the quadratic.
So $x^2-\dfrac{\sqrt5}{2}x+\dfrac14=0$.
Step 6: Clear the fractions.
Multiply by $4$: $4x^2-2\sqrt5\,x+1=0$. \[ \boxed{4x^2-2\sqrt5\,x+1=0} \]