Question

find the zeros of the quadratic polynomial 4√5x^2-24x-9√5 and verify its relationship between its zeroes and coefficients.........answer required urgently for my assignment

Answer 1

Answer:

The zeros of the quadratic polynomial are

$x=\frac{15\sqrt{5}}{10}$  and $x=-\frac{3\sqrt{5}}{10}$

The relationship between its zeroes and coefficients in the procedure

Step-by-step explanation:

step 1

Find the zeros

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$4\sqrt{5}x^{2}-24x-9\sqrt{5}=0$

so

$a=4\sqrt{5}\\b=-24\\c=-9\sqrt{5}$

substitute in the formula

$x=\frac{-(-24)(+/-)\sqrt{-24^{2}-4(4\sqrt{5})(-9\sqrt{5})}} {2(4\sqrt{5})}$

$x=\frac{24(+/-)\sqrt{-24^{2}-4(4\sqrt{5})(-9\sqrt{5})}} {8\sqrt{5}}$

$x=\frac{24(+/-)\sqrt{1,296}} {8\sqrt{5}}$

$x=\frac{24(+/-)36} {8\sqrt{5}}$

$x=\frac{24(+)36} {8\sqrt{5}}=\frac{15\sqrt{5}}{10}$

$x=\frac{24(-)36} {8\sqrt{5}}=-\frac{3\sqrt{5}}{10}$

step 2

Find the sum of the zeros and the product of the zeros

Sum of the zeros

$(\frac{15\sqrt{5}}{10})+(-\frac{3\sqrt{5}}{10})=\frac{12\sqrt{5}}{10}=\frac{6\sqrt{5}}{5}$

Product of the zeros

$(\frac{15\sqrt{5}}{10})*(-\frac{3\sqrt{5}}{10})=-\frac{9}{4}$

step 3

Verify that

Sum of the zeros= -Coefficient x/Coefficient x²

Coefficient x=-24

Coefficient x²=4√5

substitute

$\frac{6\sqrt{5}}{5}=-(-24)/4\sqrt{5}\\ \\\frac{6\sqrt{5}}{5}=\frac{6\sqrt{5}}{5}$

therefore

the relationship is verified

step 4

Verify that

Product of the zeros= Constant term/Coefficient x²

Constant term=-9√5

Coefficient x²=4√5

substitute

$-\frac{9}{4}=(-9\sqrt{5})/4\sqrt{5}\\ \\-\frac{9}{4}=-\frac{9}{4}$

therefore

the relationship is verified