Question

if K is ratio of zeros of polynomial 4 x square - 5 x minus 3 then the value of k + 1 upon K to the power minus 2 is​

Answer 1

Answer:

The required answer is in attachment.

Step-by-step explanation:

if like my writing then marks as a brain list

Answer 2

We need to recall the following formulas.

• Quadratic formula: The roots of the quadratic equation $ax^{2} +bx+c=0$ are $x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Given:

$K$ is the ratio of the zeros of a polynomial $4x^2-5x-3=0$.

Using the quadratic formula, we get

The zeros of the polynomial $4x^2-5x-3=0$ are as follows.

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

$x=\frac{5\pm\sqrt{25+48} }{8}$

$x=\frac{5\pm\sqrt{73} }{8}$

Thus, the zeros of the polynomial are $\frac{5+\sqrt{73} }{8}$ and $\frac{5-\sqrt{73} }{8}$ .

The ratio of the zeros is,

$K=\frac{\frac{5+\sqrt{73} }{8} }{\frac{5-\sqrt{73} }{8} }$

$K=\frac{5+\sqrt{73} }{5-\sqrt{73}} }$

$K=\frac{(5+\sqrt{73})(5+\sqrt{73}) }{(5-\sqrt{73})(5+\sqrt{73}) }$

$K=\frac{25+10\sqrt{73}+73 }{25-73}$

$K=\frac{98+10\sqrt{73} }{-48}$

$K=\frac{49+5\sqrt{73} }{-24}$

Now,

$K+\frac{1}{K^{-2}}=K+K^2$

$K+\frac{1}{K^{-2}}=\frac{49+5\sqrt{73} }{-24} +(\frac{49+5\sqrt{73} }{-24} )^2$

$K+\frac{1}{K^{-2}}=\frac{49+5\sqrt{73} }{-24} +\frac{2113+245\sqrt{73} }{288}$

$K+\frac{1}{K^{-2}}=\frac{1525+185\sqrt{73} }{288}$