If there is exactly one root of the equation x^2 + ax + b

Yes, you can solve it this way: \(x^2+ax+b=0\) will have only one root if it can be factored as \((x+n)^2=0\), in this case the root will be \(x=-n\). \((x+n)^2=x^2+2nx+n^2=0\) --> \(a=2n\) and \(b=n^2\). Now, since \(n=\frac{a}{2}\), then \(b=(\frac{a}{2})^2=\frac{a^2}{4}\).

Answer: E.

Or: a quadratic function is \(ax^2+bx+c=0\) will have only one root (only one intercept with x-axis) if discriminant is zero, so when \(discriminant=b^2-4ac=0\).

For given expression \(x^2+ax+b=0\) discriminant is \(a^2-4b\), so it must equal to zero: \(a^2-4b=0\) --> \(b=\frac{a^2}{4}\).

Answer: E.

Or: try number plugging, \(x^2+ax+b=0\) will have only one root if it can be factored for example as \((x+4)^2=0\) --> \(x^2+8x+16=0\) --> \(a=8\) and \(b=16\). Now, plug \(a=8\) in the answer choices and see which one gives \(b=16\), only answer choice E works.

Answer: E.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only. For example if you pick \((x+2)^2=0\) then you get two "correct" options B and E.

Hope it helps.