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:[tex]y=x^{2} -\frac{48}{7}x+\frac{64}{7}[/tex]Step-by-step explanation:step 1Find the roots of the quadratic equationwe know that The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to [tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex] in this problem we have [tex]x^{2} -6x+7=0[/tex]  so [tex]a=1\\b=-6\\c=7[/tex] substitute in the formula [tex]x=\frac{6(+/-)\sqrt{-6^{2}-4(1)(7)}} {2(1)}[/tex] [tex]x=\frac{6(+/-)2\sqrt{2}} {2}[/tex] [tex]x=3(+/-)\sqrt{2}[/tex][tex]x1=3(+)\sqrt{2}[/tex][tex]x2=3(-)\sqrt{2}[/tex]The roots of the equation are a and bso[tex]a=3(+)\sqrt{2}[/tex][tex]b=3(-)\sqrt{2}[/tex]step 2Find a quadratic equation with roots a+1/b and b + 1/aso[tex]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})[/tex][tex]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})[/tex]The quadratic equation is equal to[tex]y=(x-\frac{8}{7}(3+\sqrt{2}))(x-\frac{8}{7}(3-\sqrt{2}))[/tex][tex]y=x^{2} -\frac{8}{7}(3-\sqrt{2})x-\frac{8}{7}(3+\sqrt{2})x+\frac{64}{49}(7)[/tex][tex]y=x^{2} -\frac{48}{7}x+\frac{64}{7}[/tex]