Find the formula, in standard form $y=a x^{2}+b x+c$, for a quadratic that has roots at $x=\sqrt{26}$ and $x=-\sqrt{26}$, and has leading coefficient of 1 .
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y=
\]
Answer
\(\boxed{y = x^{2} - 26}\) is the standard form of the quadratic equation.
Step 1 ：The roots of a quadratic equation are given by the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this case, we know that the roots are \(x=\sqrt{26}\) and \(x=-\sqrt{26}\), and the leading coefficient (a) is 1. We can use this information to find the values of b and c.
Step 2 ：Since the roots are \(\sqrt{26}\) and \(-\sqrt{26}\), we know that the quadratic equation can be written in the form \(y = (x - \sqrt{26})(x + \sqrt{26})\). We can expand this to find the standard form of the equation.
Step 3 ：The expanded form of the equation is \(y = x^{2} - 26\). This is already in standard form, with \(a = 1\), \(b = 0\), and \(c = -26\).
Step 4 ：\(\boxed{y = x^{2} - 26}\) is the standard form of the quadratic equation.
