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Discussion
We've previously learned that the solutions to a quadratic equation, ax2+bx+c=0 can always be found using the quadratic formula:
But what if a parabola doesn't intersect the x-axis at all? What if it intersects only once? Examine the following parabola:
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How many real solutions do you see?
The number of real solutions is the same as the number of times the parabola meets the x-axis.
From the graph we see it does not intersect the x-axis at all, so there are no real roots.
The graph above can be represented by y=x2+1. We know that the quadratic formula should always yield an answer, so where's the disconnect? Let's try to solve for the roots. With y=x2+1, we can see that a=1,b=0, and c=1. Substituting into the formula:
we run into √−4, which has no real solutions (since we can't take the square root of a negative number). Later on, we'll learn about something called imaginary numbers which enable us to find solutions. But for now, this isn't something we can solve directly.
This makes sense: there are no intersections with the graph on the x-axis, so we wouldn't expect to find a real x-value from solving.
Discussion
Until now, we've relied on techniques like factoring or using the quadratic formula to explicitly find solutions to quadratic equations. But our example of y=x2+1 raises a more basic question:
Can we tell, just by examining the values of a,b, and c, whether real roots exist?
The answer is yes!
Recall that the only obstacle in the quadratic formula had to do with taking the square root of a negative number. Let's examine this expression under the square root more closely:
This expression appears in every quadratic we solve. It's sign (positive, negative, or zero) controls whether the square root (and therefore the roots themselves) are real or not.
We call this expression the discriminant because it well, discriminates among the three possible root counts in a single expression.
The discriminant and solution count
The discriminant of a quadratic is the term under the square root in the quadratic formula:
When Δ<0, the square root has a negative value inside, and so the quadratic has no real solutions.
When Δ=0, the square root is zero, and the ±√Δ makes no difference, so there is only one real solution.
When Δ>0, √Δ is positive and so ±√Δ yields two real roots.
Exercise
For what value(s) of k does the quadratic 2x2−3x+k have 2 real roots?
Select the correct option
Discussion
We just saw that the discriminant tells us how many real roots a quadratic has, without even solving for them directly.
Interestingly enough, there's another powerful shortcut: we can determine the sum and product of the roots themselves directly from the coefficients of a quadratic.
Think back to factored form. We know a quadratic can be written as y=a(x−α)(x−β), where α and β are the x-intercepts (solutions).
Try expanding the right side of that equation, and compare your result with the general form, ax2+bx+c.
What relationships do you notice between the roots α,β and the coefficients a,b, and c?
By expanding
we notice that the sum α+β and product αβ of the roots directly impact the general form of the quadratic.
Matching coefficients with y=ax2+bx+c:
Hence the sum of the roots is α+β=−ab and their product is αβ=ac. These are known as Vieta's Formulas.
Vieta's Formulas
In a quadratic ax2+bx+c, the roots α and β satisfy
Checkpoint
Find the sum and product of the roots of 5x2−10x+3.
Select the correct option
Discussion
Up to now we have treated the roots of a quadratic as significant points where the graph crosses (or touches) the x-axis.
But for an inequality, say ax2+bx+c>0 the goal shifts: we want to find every x-value where the curve lies above the x-axis, rather than just where it crosses it.
Think about what affects the interval where the inequality holds true.
What characteristics of a quadratic do you think affect the answer?
Two elements of a quadratic inequality control the answer:
The roots—they mark the exact x-values where the expression switches sign.
The concavity—we know that the sign of the leading coefficient a tells you whether the arms of the parabola point upward or downward. This can inform us which side of each root lies above (>0) or below (<0) the axis.
Once the roots are marked on a number line, those points split the line into separate intervals.
Because a quadratic can switch sign only at its roots, every interval is either entirely positive or entirely negative. You have two quick ways to decide which is which:
Test-point method: Pick one convenient value inside each interval, substitute it into the quadratic, and record the sign. Whatever sign you get applies to the whole interval.
Concavity shortcut: Notice that the sign pattern must alternate as you cross each root.
If the parabola opens upward (a>0), the two outer intervals (far left and far right) are positive, and the interior intervals alternate accordingly.
If it opens downward (a<0), the outer intervals are negative, with the signs alternating in the same way.
Additionally, note that inequalities can be strict (<,>) or inclusive (≤,≥), which dictate if we should include the roots themselves.
Quadratic Inequalities
A quadratic inequality is an inequality of the form
They can be solved by finding the roots of the quadratic and the concavity of the parabola.
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Checkpoint
Solve the inequality −x2+2x+3>0.
Select the correct option
Frequently, you will be asked to apply quadratics to real-world situations. You might be asked to find the vertex of some given parabola, and explain its significance in a physical context.
For example, suppose a ball is thrown into the air and its height over time is modeled by a quadratic equation. The vertex of that parabola represents the ball’s highest point — the peak of its flight. Similarly, in business or economics, a quadratic might represent profit over time or revenue as a function of price. In that case, the vertex often tells you the maximum profit or revenue possible.
You might also have to find the roots and use factored form. For example:
In a projectile motion scenario, the roots might represent when an object takes off and when it hits the ground.
In a business context, they could represent break-even points — the input values that result in zero profit.
There are many other cases where you could encounter similar situations. Be sure to keep this in mind!
One last note! Sometimes, you might encounter a quadratic in hiding—a pseudo-quadratic.
Frequently, you'll still be able to factor using the methods discussed in Foundations of Quadratics. Some common examples can involve factoring with ex. Let's look at an example to visualize this idea:
Factor e2x−5ex+6.
Let u=ex. Then the equation becomes
which factors as
Hence u=2 or u=3. Substituting back gives
Therefore the solutions are x=ln(2) and x=ln(3).
Another common case of substitution involves expressions such as x4 (where you would substitute with x2). The strategy is always to think about what you can substitute, then use traditional factoring methods, then substitute back to get a final answer.
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