Quadratic Parameters and Root Relationships, SAT math lesson

Some SAT questions hand you an equation with an extra letter in it, such as $x^2 - 6x + k = 0$ , and tell you something about its solutions. Maybe it has exactly one solution, or one solution is $2$ , or the solutions add to a certain number. Your job is to run the usual reasoning backwards and find the letter. The tools are ones you already know, the discriminant and the link between roots and coefficients. The new skill is keeping track of which letter is which.

See also The Quadratic Formula and Real Solutions in Foundations for Calculus and Quadratic Functions and Vertex Form in Precalculus.

Key terms

Parameter

Definition A parameter is a letter in an equation that stands for a fixed but unknown constant, as opposed to the variable, which is the letter the equation is solved for. In $x^2 - 6x + k = 0$ the variable is $x$ and $k$ is a parameter. Each value of $k$ gives a different equation, with its own solutions in $x$ .

In plain words First you find the number the parameter must be. Then, with that number plugged in, you solve for the variable. These are two different answers, and a question asks for one of them.

Example The equation $x^2 - 6x + k = 0$ has $x = 2$ as a solution. Substituting $x = 2$ gives $4 - 12 + k = 0$ , so the parameter is $k = 8$ . With $k = 8$ the equation is $x^2 - 6x + 8 = 0$ , which factors as $(x - 2)(x - 4) = 0$ . So the solutions are $x = 2$ and $x = 4$ . If a question asks for the other solution, the answer is $4$ , not $8$ .

Discriminant condition

Definition For $ax^2 + bx + c = 0$ with real coefficients and $a \ne 0$ , the discriminant $D = b^2 - 4ac$ decides the number of real solutions.

Two distinct real solutions exactly when $D > 0$ .
Exactly one real solution exactly when $D = 0$ .
No real solutions exactly when $D < 0$ .

When $a$ , $b$ or $c$ contains a parameter, each of these becomes an equation or inequality for the parameter. The condition is only valid while $a \ne 0$ .

In plain words Translate the words about solutions into a statement about $D$ , then solve that statement for the parameter.

Example For which values of $k$ does $x^2 - 6x + k = 0$ have no real solutions? Here $a = 1$ , $b = -6$ and $c = k$ , so

\begin{gathered} D = 36 - 4k < 0 \\[0.4em] \text{exactly when} \quad k > 9. \end{gathered}

There is a picture behind this. The equation is the same as $x^2 - 6x = -k$ , so its solutions are where the parabola $y = x^2 - 6x$ meets the horizontal line $y = -k$ . The parabola’s lowest point is $(3, -9)$ . A line above that height crosses it twice, the line $y = -9$ touches it once, and a lower line misses it. In the Desmos graphing calculator you can graph $y = x^2 - 6x$ and then type $y = -k$ . Since $k$ is a new letter, Desmos offers to add a slider for it. Add the slider and drag it to see all three cases.

Repeated root

Definition A quadratic equation with $a \ne 0$ has a repeated root exactly when $D = 0$ . The equation is then $a(x - r)^2 = 0$ , and its one real solution is

r = -\frac{b}{2a}.

The phrases “exactly one real solution” and “the graph touches the $x$ -axis at one point” both describe this case.

In plain words Set the discriminant equal to zero and solve for the parameter first. Only then find the root, using the parameter value you got.

Example For which values of $b$ does $x^2 + bx + 16 = 0$ have exactly one real solution? The condition is $b^2 - 64 = 0$ , so $b = 8$ or $b = -8$ . Each value gives its own repeated root.

$b$	Equation	Repeated root
$8$	$(x + 4)^2 = 0$	$x = -4$
$-8$	$(x - 4)^2 = 0$	$x = 4$

Forgetting $b = -8$ is the usual mistake, since $b^2 = 64$ has two solutions.

Sum of roots

Definition If $a \ne 0$ and $ax^2 + bx + c = 0$ has real solutions $r$ and $s$ (with $r = s$ for a repeated root), then

r + s = -\frac{b}{a}.

The reason is that the quadratic equals $a(x - r)(x - s) = ax^2 - a(r + s)x + ars$ , and matching the $x$ coefficients gives $b = -a(r + s)$ . The equation must be in standard form, with every term on one side and zero on the other, before you read off $a$ and $b$ . When $D < 0$ there are no real solutions to add, although the same identity holds for the two complex solutions.

In plain words The two roots always add up to the opposite of $b$ divided by $a$ . You can find that sum without finding the roots.

Example For $2x^2 - 7x + 3 = 0$ the sum of the solutions is

-\frac{-7}{2} = \frac{7}{2}.

As a check, the solutions are $3$ and $\frac{1}{2}$ , and $3 + \frac{1}{2} = \frac{7}{2}$ .

Product of roots

Definition Under the same conditions, $a \ne 0$ and real solutions $r$ and $s$ ,

rs = \frac{c}{a}.

As with the sum, it also holds for the two complex solutions when $D < 0$ . It comes from matching the constant terms in $a(x - r)(x - s) = ax^2 + bx + c$ , which gives $ars = c$ .

In plain words The product of the roots is the constant term divided by the leading coefficient, once the equation is in standard form.

Example The equation $3x^2 = 12 - 5x$ is not in standard form yet. Moving every term to the left gives $3x^2 + 5x - 12 = 0$ , so the product of the solutions is

\frac{-12}{3} = -4,

and their sum is $-\frac{5}{3}$ . The actual solutions are $\frac{4}{3}$ and $-3$ , and indeed $\frac{4}{3} \cdot (-3) = -4$ . Reading $c = 12$ from the original equation would have given the wrong sign.

Parameter cases

Definition When the parameter appears in the coefficient of $x^2$ , the equation is quadratic only for the parameter values that keep that coefficient nonzero. Solve it in cases. For the values that make the leading coefficient zero, solve the resulting linear (or constant) equation directly. For all other values, use the discriminant, the quadratic formula or the root relationships. Then combine the results.

In plain words Before using any quadratic rule, ask whether some value of the parameter wipes out the $x^2$ term. If it does, that value needs its own check.

Example For which values of $k$ does $(k - 2)x^2 + 4x + 2 = 0$ have exactly one real solution?

When $k = 2$ the equation is $4x + 2 = 0$ , which has exactly one solution, $x = -\frac{1}{2}$ . So $k = 2$ works.
When $k \ne 2$ the equation is quadratic, and exactly one real solution means $D = 0$ .

\begin{aligned} D &= 16 - 4(k - 2)(2) \\[0.4em] &= 32 - 8k = 0, \end{aligned}

so $k = 4$ . Then the equation is $2x^2 + 4x + 2 = 2(x + 1)^2 = 0$ , with the repeated root $x = -1$ . Both $k = 2$ and $k = 4$ work. Using only the discriminant would have missed $k = 2$ . The discriminant test does not apply there, because the equation is not quadratic.

Check yourself

In the equation $x^2 + 10x + c = 0$ , $c$ is a constant. If the equation has exactly one real solution, what is the value of $c$ ? (A) $5$ , (B) $10$ , (C) $25$ , (D) $100$

Show the answer

Choice (C) is correct. Exactly one real solution means

D = 10^2 - 4c = 0,

so $c = 25$ , and the equation is $(x + 5)^2 = 0$ . Choice (D) is $b^2$ without dividing by $4$ . Choice (A) is half of $b$ , which is the repeated root’s size, not the constant term, and choice (B) just repeats $b$ .

One solution of the equation $2x^2 + bx - 15 = 0$ is $x = 3$ , where $b$ is a constant. What is the other solution?

Show the answer

The answer is $-\frac{5}{2}$ , which you can enter as $-5/2$ or $-2.5$ . The product of the solutions is

\frac{-15}{2},

so the other solution is $-\frac{15}{2} \div 3 = -\frac{5}{2}$ . You can confirm it another way. Substituting $x = 3$ gives $18 + 3b - 15 = 0$ , so $b = -1$ , and $2x^2 - x - 15 = (2x + 5)(x - 3)$ . Note that $b = -1$ is the parameter value, not the solution asked for.

In the equation $kx^2 - 6x + 3 = 0$ , $k$ is a constant. For which values of $k$ does the equation have exactly one real solution? (A) $3$ only, (B) $0$ only, (C) $0$ and $3$ , (D) $-3$ and $3$

Show the answer

Choice (C) is correct. When $k = 0$ the equation is $-6x + 3 = 0$ , which has the single solution $x = \frac{1}{2}$ . When $k \ne 0$ it is quadratic, and $D = 36 - 12k = 0$ gives $k = 3$ , with the repeated root $x = 1$ . Choice (A) forgets the linear case, choice (B) forgets the discriminant case, and choice (D) treats $k = 3$ like a square root that needs a negative partner.

Quadratic Parameters and Root Relationships

Theory, part 55

Practice, part 56

Key terms

Parameter

Discriminant condition

Repeated root

Sum of roots

Product of roots

Parameter cases

Check yourself

How this lesson connects

Builds on

Used later in