In the equation above, k is a constant. For which of the following values f k does the equation have at least one re3al solution ?

Options

A. 2
B. 3
C. 4
D. 5

Answer

A

Explanation

Restate the equation in "standard" quadratic form, `3x ^(2) -5x +k =0.` Recall that a quadratic equation has two real solutions if the discriminat, `b ^(2)-4ac,` is positive. The equation has one real soluation if the discriminant is 0. For the given equation, `a =3, b=-5, anad c=k,` so you can set up the inequality `(-5)^(2)-4 (3)k ge 0.` This simplifies to `25 ge 12k.` The only choice for which this is true is (A) if you didn't recall teh shortcut about the siscriminant, consider the quadratic formula, `x = (-b pmsqrt(b ^(2) -4ac))/(2a)` If the term inside the radical, `b ^(2), 4ac,` is negative, then there are no real solutions.