Answer:
[tex]\textsf{Zeros}: \quad x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}[/tex]
Step-by-step explanation:
Rewrite the given polynomial in the form ax² + bx + c:
[tex]f(x)=2 \sqrt{3}x^2+5x-4 \sqrt{3}[/tex]
To find the zeros, set the function to zero and solve for x using the quadratic formula.
[tex]\implies 2 \sqrt{3}x^2+5x-4 \sqrt{3}=0[/tex]
Quadratic formula:
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
Therefore,
Substituting the values into the quadratic formula:
[tex]\implies x=\dfrac{-5 \pm \sqrt{5^2-4(2\sqrt{3})(-4\sqrt{3})} }{2(2\sqrt{3})}[/tex]
[tex]\implies x=\dfrac {-5 \pm \sqrt {121}}{4\sqrt{3}}[/tex]
[tex]\implies x=\dfrac {-5 \pm 11}{4\sqrt{3}}[/tex]
[tex]\implies x=\dfrac {6}{4\sqrt{3}}, \:\:x=\dfrac {-16}{4\sqrt{3}}[/tex]
[tex]\implies x=\dfrac {3}{2\sqrt{3}}, \:\:x=-\dfrac {4}{\sqrt{3}}[/tex]
[tex]\implies x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}[/tex]
The sum of the roots of a polynomial is -b/a:
[tex]\implies -\dfrac{b}{a}=-\dfrac{5}{2 \sqrt{3}}=-\dfrac{5\sqrt{3}}{6}[/tex]
The sum of the found roots is:
[tex]\implies \left(\dfrac {\sqrt{3}}{2}\right)+\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{5\sqrt{3}}{6}[/tex]
Hence proving the sum of the roots is -b/a
The product of the roots of a polynomial is: c/a
[tex]\implies \dfrac{c}{a}=\dfrac{-4\sqrt{3}}{2\sqrt{3}}=-2[/tex]
The product of the found roots is:
[tex]\implies \left(\dfrac {\sqrt{3}}{2}\right)\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{12}{6}=-2[/tex]
Hence proving the product of the roots is c/a
Therefore, the relationship between the roots and the coefficients is verified.
