Finding a Quadratic’s Linear Term


In the equation \fn_jvn \small 4x^2+bx+c=0, b and c are integers and there is only one real root. Is \fn_jvn b^2 a multiple of 4?


This question is about quadratic equations. You can solve some quadratic equations by factoring, but not this one, because it has too many unknowns. Another way to solve quadratic equations is to use the quadratic formula. For a quadratic equation \fn_jvn ax^2+bx+c=0, the quadratic formula says

\fn_jvn x=\frac{-b\pm \sqrt{b^2-4ac} }{2a}

For the given problem, \fn_jvn \small a=4 and b and c are, well, b and c. Substitute the one value you are given, \fn_jvn \small a=4, into the quadratic formula and see what happens.

\fn_jvn \small x=\frac{-b\pm \sqrt{b^2-4 \cdot 4c}}{2 \cdot 4}=\frac{-b\pm \sqrt{b^2-16c}}{8}

The problem tells you there is only one solution. That means the positive and negative values of the radical must be the same — which means they must equal zero — which means the radicand, \fn_jvn \small b^2-4ac, must equal zero. And that tells you something about the relationship between b and c:

\fn_jvn \small b^2-16c=0

\fn_jvn \small b^2=16c

\fn_jvn \small b=\pm 4\sqrt{c}

The problem tells you b and c are integers. If b is an integer and \fn_jvn \small b=\pm 4\sqrt{c}, then \fn_jvn \small \sqrt{c} must be an integer, so c must be a perfect square. Then \fn_jvn \small \pm 4\sqrt{c} contains a factor of 4, and since \fn_jvn \small \pm 4\sqrt{c} equals \fn_jvn \small b^2, \fn_jvn \small b^2 must also contain a factor of 4 — that is, \fn_jvn \small b^2 is a multiple of 4.