Question

$root7y^2 - 6y - 13root7 = 0$

Answer 1

$\bold{\blue{\underline{\red{G}\pink{iv}\green{en}\purple{:-}}}}$

• √7y² - 6y - 13√7 = 0

$\bold{\blue{\underline{\red{Zeros}\pink{\:of\:}\green{\:the\:Quadratic\:equation}\purple{:-}}}}$

• Zeros of the quadratic equation is the value of the variable for which when replaced in the quadratic equation the value for the quadratic equation changes to zero.

$\bold{\blue{\underline{\red{General}\pink{\:form \:of }\green{\:Quadratic\:equation}\purple{:-}}}}$

• ax² + bx + c = 0

$\bold{\blue{\underline{\red{Q}\pink{uest}\green{ion}\purple{:-}}}}$

• To find the zeros of the quadratic equation .

$\bold{\blue{\underline{\red{S}\pink{olut}\green{ion}\purple{:-}}}}$

• a = √7
• b = -6
• c = 13√7

• Quadratic formula =  $\bold{\frac{-b\: \pm\:{\sqrt{b^{2}\:-\:4ac} }}{2a}}$

• x =  $\bold{\frac{-(-6)\: \pm\:{\sqrt{-6^{2}\:-\:4(\sqrt{7})(-13\sqrt{7})} }}{2\sqrt{7}}}$

• x = $\bold{\frac{6\: \pm\:\sqrt{36\:-\:(4)(-91)}}{2\sqrt{7}}}$  

• x = $\bold{\frac{6\: \pm\:\sqrt{36\:+\:364}}{2\sqrt{7}}}$

• x =  $\bold{\frac{6\: \pm\:\sqrt{400}}{2\sqrt{7}}}$

• x = $\bold{\frac{6\: \pm\:20}{2\sqrt{7}}}$

• x = $\bold{\frac{6\:-\:20}{2\sqrt{7}}}$

• x = $\bold{\frac{-14}{2\sqrt{7}}}$  

• x = $\bold{\frac{-7}{\sqrt{7}}}$

• x = $\bold{\frac{-7\:\times\:\sqrt{7} }{\sqrt{7}\:\times\:\sqrt{7} }}$

• x = $\bold{\frac{-7\:\times\:\sqrt{7} }{7}}$

   

       $\bold{\boxed{\red{x = -\sqrt{7}}}}$

• x = $\bold{\frac{6\:+\:20}{2\sqrt{7}}}$

• x = $\bold{\frac{26}{2\sqrt{7}}}$

• x = $\bold{\frac{13}{\sqrt{7}}}$  

• x = $\bold{\frac{13\:\times\:\sqrt{7} }{\sqrt{7}\:\times\:\sqrt{7} }}$

        $\bold{\boxed{\red{\frac{13\sqrt{7} }{7}}}}$