Question

find the zeros of following quadratic polynomial and verify the relationship between the zeros and the coefficient of the polynomial f(x)=5x-4√3+2√3x²​

Answer 1

Answer:

$\textsf{Zeros}: \quad x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}$

Step-by-step explanation:

Rewrite the given polynomial in the form ax² + bx + c:

$f(x)=2 \sqrt{3}x^2+5x-4 \sqrt{3}$

To find the zeros, set the function to zero and solve for x using the quadratic formula.

$\implies 2 \sqrt{3}x^2+5x-4 \sqrt{3}=0$

Quadratic formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore,

• a = 2√3
• b = 5
• c = - 4√3

Substituting the values into the quadratic formula:

$\implies x=\dfrac{-5 \pm \sqrt{5^2-4(2\sqrt{3})(-4\sqrt{3})} }{2(2\sqrt{3})}$

$\implies x=\dfrac {-5 \pm \sqrt {121}}{4\sqrt{3}}$

$\implies x=\dfrac {-5 \pm 11}{4\sqrt{3}}$

$\implies x=\dfrac {6}{4\sqrt{3}}, \:\:x=\dfrac {-16}{4\sqrt{3}}$

$\implies x=\dfrac {3}{2\sqrt{3}}, \:\:x=-\dfrac {4}{\sqrt{3}}$

$\implies x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}$

The sum of the roots of a polynomial is -b/a:

$\implies -\dfrac{b}{a}=-\dfrac{5}{2 \sqrt{3}}=-\dfrac{5\sqrt{3}}{6}$

The sum of the found roots is:

$\implies \left(\dfrac {\sqrt{3}}{2}\right)+\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{5\sqrt{3}}{6}$

Hence proving the sum of the roots is -b/a

The product of the roots of a polynomial is:  c/a

$\implies \dfrac{c}{a}=\dfrac{-4\sqrt{3}}{2\sqrt{3}}=-2$

The product of the found roots is:

$\implies \left(\dfrac {\sqrt{3}}{2}\right)\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{12}{6}=-2$

Hence proving the product of the roots is c/a

Therefore, the relationship between the roots and the coefficients is verified.