The expression 8x² + 4x + n has real roots if the discriminant is non-negative. For the quadratic equation ax² + bx + c, the discriminant is calculated as b² - 4ac. In this case, substituting a = 8, b = 4, and c = n leads to the condition 4² - 4(8)(n) ≥ 0. Solving this inequality reveals that n must be less than or equal to 0.5 for the expression to have real roots.
PREREQUISITESStudents studying algebra, educators teaching quadratic equations, and anyone interested in understanding the conditions for real roots in polynomial expressions.
MarkFL said:Consider the general quadratic:
$$ax^2+bx+c$$
If this quadratic is to have real roots, then the discriminant has to be non-negative, that is:
$$b^2-4ac\ge0$$
What do you get when you apply this rule to the quadratic you posted?
RTCNTC said:Are you talking about the discriminant?
MarkFL said:Yes, the discriminant is what's under the radical in the quadratic formula, and it cannot be negative if the roots ore real. :D
RTCNTC said:I plug a = 8, b = 4 and c = n into b^2 - 4ac and equate to 0, right?
MarkFL said:You want to set the discriminant greater than or equal to zero, since you don't want it to be negative. If the discriminant is equal to zero, then you will have a repeated root, or a root of multiplicity 2. If the discriminant is greater than zero, then you will have two distinct roots.
To put this in graphical terms, when the discriminant is zero, then the parabola will be tangent to the $x$-axis, that is, the parabola will only touch the $x$-axis at one point. If the discriminant is greater than zero, then the parabola will cross the $x$-axis at two points, and if the discriminant is negative then the parabola will not touch the $x$-axis at all.