Question

the roots of the equation x^2-6x+7=0 are a and b.
find a quadratic equation with roots a+1/b and b + 1/a.

please reply ASAP as i have an exam tomorrow!!​

Answer 1

Answer:

$y=x^{2} -\frac{48}{7}x+\frac{64}{7}$

Step-by-step explanation:

step 1

Find the roots of the quadratic equation

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

$x^{2} -6x+7=0$  

so

$a=1\\b=-6\\c=7$

substitute in the formula

$x=\frac{6(+/-)\sqrt{-6^{2}-4(1)(7)}} {2(1)}$

$x=\frac{6(+/-)2\sqrt{2}} {2}$

$x=3(+/-)\sqrt{2}$

$x1=3(+)\sqrt{2}$

$x2=3(-)\sqrt{2}$

The roots of the equation are a and b

so

$a=3(+)\sqrt{2}$

$b=3(-)\sqrt{2}$

step 2

Find a quadratic equation with roots a+1/b and b + 1/a

so

$a+\frac{1}{b} =(3+\sqrt{2})+\frac{1}{3-\sqrt{2}} =\frac{9-2+1}{3-\sqrt{2}} =\frac{8}{3-\sqrt{2}}=\frac{8}{3-\sqrt{2}}*(\frac{3+\sqrt{2}}{3+\sqrt{2}})=\frac{8}{7}(3+\sqrt{2})$

$b+\frac{1}{a} =(3-\sqrt{2})+\frac{1}{3+\sqrt{2}} =\frac{9-2+1}{3+\sqrt{2}} =\frac{8}{3+\sqrt{2}}=\frac{8}{3+\sqrt{2}}*(\frac{3-\sqrt{2}}{3-\sqrt{2}})=\frac{8}{7}(3-\sqrt{2})$

The quadratic equation is equal to

$y=(x-\frac{8}{7}(3+\sqrt{2}))(x-\frac{8}{7}(3-\sqrt{2}))$

$y=x^{2} -\frac{8}{7}(3-\sqrt{2})x-\frac{8}{7}(3+\sqrt{2})x+\frac{64}{49}(7)$

$y=x^{2} -\frac{48}{7}x+\frac{64}{7}$