Showing a Condition Holds for Every Value

Some questions ask you to show that a result holds for all real values of a constant \(k\): the roots of an equation are always real, or never real. The discriminant \(D=b^2-4ac\) (also written \(\Delta\)) is then an expression in \(k\), and you must find its sign without knowing \(k\).

Why: a square is never negative. \((k-2)^2\ge0\) for every real \(k\), and it is zero only when \(k=2\), so its graph touches the \(k\)-axis there. Adding a positive number lifts the graph clear of the axis: \((k-2)^2+4\ge4>0\). Once \(D\) is written as a square, or as a square plus a positive number, its sign is settled for every \(k\).

Method

1. Write the equation as \(ax^2+bx+c=0\).
2. If the constant appears in \(a\), substitute the value that makes \(a=0\) and solve what is left separately.
3. Write \(D=b^2-4ac\) and simplify it to an expression in the constant.
4. Write \(D\) as a perfect square, such as \(4(k-1)^2\), or complete the square, such as \((k-3)^2+5\).
5. State the sign of \(D\) for every value of the constant and what it means, then finish with the sentence the question asks for, for example "the equation has real roots for all real values of \(k\)".

Example

Show that \(x^2+kx+k-1=0\) has real roots for all real values of \(k\).

\(\begin{aligned} D&=k^2-4(1)(k-1) \\ &=k^2-4k+4 \\ &=(k-2)^2\ge0 \end{aligned}\)

Since \(D\ge0\) for every real \(k\), the equation has real roots for all real values of \(k\).

When the constant multiplies \(x^2\), check the zero case first. For \(kx^2-(k+1)x+1=0\): if \(k=0\), the equation is \(-x+1=0\), so \(x=1\) is a real root; if \(k\ne0\), \(D=(k+1)^2-4k=(k-1)^2\ge0\). The roots are real for all real values of \(k\).

Exam check "Real roots" needs only \(D\ge0\), so a perfect square is enough. "Two distinct real roots" needs \(D>0\): \((k-2)^2\) is not enough, because it is zero when \(k=2\), but \((k-2)^2+4\) is. "No real roots" needs \(D<0\) for every \(k\), for example \(D=-4(k^2+1)\). Never substitute a few values of \(k\): that does not show the result for all \(k\).