# Not sure how to answer this please help. The parabola y=x2 + kx + 1 has exactly one real root. What are the possible values for k?

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A quadratic function in one variable ( as in your problem where `y=f(x)=x^2+kx+1) ` has exactly two roots in the complex numbers according to the fundamental theorem of algebra.

There are three possible cases:

(1) Both of the roots are real. In the graph, there are two places where the graph intersects the x-axis.

(2) Both of the roots are imaginary. In the graph, there are no intersections with the x-axis. (The graph opens up and is entirely above the x-axis or the graph opens down and is entirely below the x-axis.

(3) There is exactly on real root. The graph is tangent to the x-axis. This root is called a double root or a repeated root. There are two roots, but they are the same number.

To determine which case is present, we use the determinant. If the quadratic is written in standard form `y=ax^2+bx+c ` then the determinant is `b^2-4ac ` . You may recognize this expression as the expression in teh radicand of the quadratic formula.

If the determinant is positive, there are two real roots. If the determinant is a perfect square, the zeros are rational.)

If the determinant is negative there are no real zeros -- both are imaginary.

If the determinant is zero, there is exactly one real root.

(Consider the quadratic formula `x=(-b +- sqrt(b^2-4ac))/(2a) ` . The vertical line of symmetry is `x=(-b)/(2a) ` , and when the determinant is positive the zeros are symmetrically placed about this line. If the determinant is zero, the two zeros are both on the line of symmetry and must be the vertex.)

For your problem, set the discriminant equal to zero.

`k^2-4(1)(1)=0 `

Then `k^2=4 ==> k=+-2 `

When k=2, the function factors as y=(x+1)(x+1) and for k=-2 it factors as y=(x-1)(x-1); both of these have double roots.

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k is either 2 or -2

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