Answer:
To solve the given quadratic equation solve using completing the perfect square
The two zeros of quadratic equation \(x^2-8x=242\) are \(x=\sqrt{258}+4,\:x=-\sqrt{258}+4\)
Step-by-step explanation:
Given quadratic equation \(x^2-8x=242\)
We have to choose the most appropriate strategy to solve the given quadratic equation \(x^2-8x=242\).
The given quadratic equation \(x^2-8x=242\)
Solve using completing the perfect square, \(\mathrm{Write\:equation\:in\:the\:form:\:\:}x^2+2ax+a^2=\left(x+a\right)^2\)
Solve for a ,
\(2ax=-8x\) , we get a = -4
\(\mathrm{Add\:}a^2=\left(-4\right)^2\mathrm{\:to\:both\:sides}\)
\(x^2-8x+\left(-4\right)^2=242+\left(-4\right)^2\)
left side becomes a perfect square, we get,
\(\left(x-4\right)^2=258\)
\(\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}\)
\(x-4=\pm\sqrt{258}\\\\\mathrm{Solve\:}\:x-4=\sqrt{258}:\quad x=\sqrt{258}+4\)
\(\mathrm{Solve\:}\:x-4=-\sqrt{258}:\quad x=-\sqrt{258}+4\)
Thus, the two zeros of quadratic equation \(x^2-8x=242\) are \(x=\sqrt{258}+4,\:x=-\sqrt{258}+4\)
