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The Quadratic Formula and the Discriminant: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A rearranged equation

Solve $2x(x+1)=x+4$ over the real numbers, giving exact values.
1. Hint 1
  The coefficients belong to the equation after everything has been moved to one side.
2. Hint 2
  Use the resulting quadratic’s coefficients and keep the radical exact.
Answer

$x=\frac{-1\pm\sqrt{33}}4$ .

Full solution

The standard equation is $2x^2+x-4=0$ , so $a=2$ , $b=1$ , $c=-4$ .
$\begin{aligned} \Delta&=1-4(2)(-4) \\ \Delta&=33 \\ x&=\frac{-1\pm\sqrt{33}}4 \end{aligned}$

For either root, $4x+1=\pm\sqrt{33}$ ; squaring gives $16x^2+8x=32$ , equivalent to the original equation.

Answer

$x=\frac{-1\pm\sqrt{33}}4$ .

Key idea

Identify coefficients after collecting the equation, then apply the formula to all of them.
Problem 1: Mark complete
Problem 2 Counting an output level

How many real inputs make $f(x)=2x^2-4x+7$ equal $6$ ?
1. Hint 1
  Root counts apply to the expression after subtracting the requested output.
2. Hint 2
  Calculate the discriminant of that shifted quadratic.
Answer

Two real inputs.

Full solution

The relevant equation is
$2x^2-4x+1=0.$
Its discriminant is
$\begin{aligned} \Delta&=(-4)^2-4(2)(1) \\ \Delta&=8 \end{aligned}$

The positive discriminant gives two distinct real roots, hence two inputs at the requested height.

Answer

Two real inputs.

Key idea

To count inputs at a specified output, compute the discriminant after subtracting that output.
Problem 2: Mark complete
Problem 3 A rational factor question

Determine whether $4x^2+3x-2$ factors into linear factors with rational coefficients.
1. Hint 1
  Confirm the coefficients are integers, then compute the discriminant.
2. Hint 2
  Use the discriminant to decide whether its square root is rational.
Answer

No, it does not factor into linear factors with rational coefficients.

Full solution

The expression has integer coefficients.
$\begin{aligned} \Delta&=3^2-4(4)(-2) \\ \Delta&=41 \end{aligned}$

Since $41>0$ , the equation has two real roots, but $41$ is not a perfect square, so those roots are irrational.
A quadratic with integer coefficients factors into linear factors with rational coefficients only when its discriminant is a nonnegative perfect square, so this one does not, even though it does have real roots.

Answer

No, it does not factor into linear factors with rational coefficients.

Key idea

For quadratics with integer coefficients, a positive discriminant gives two real roots, which are rational exactly when the discriminant is a perfect square.
Problem 3: Mark complete
Problem 4 A growing measurement

A device’s reading after $t\ge0$ seconds is $R(t)=3t^2+2t+1$ . At what time does it first read $4$ ? Give the exact time.
1. Hint 1
  Turn the target reading into a quadratic equation.
2. Hint 2
  Keep only roots in the stated time domain.
Answer

$t=\frac{\sqrt{10}-1}{3}$ seconds.

Full solution

The equation is $3t^2+2t-3=0$ .
$\begin{aligned} \Delta&=4+36 \\ \Delta&=40 \\ t&=\frac{-2\pm\sqrt{40}}6 \\ t&=\frac{-1\pm\sqrt{10}}3 \end{aligned}$

The minus choice is negative and is excluded.
The plus choice is positive, and since the other algebraic root is outside the domain, this is the unique allowed time.
Substituting $t=\tfrac{\sqrt{10}-1}3$ gives $3t^2+2t=3$ , so $R(t)=3+1=4$ , confirming the check.

Answer

$t=\frac{\sqrt{10}-1}{3}$ seconds.

Key idea

The formula supplies algebraic roots, and a model’s domain selects the meaningful one.
Problem 4: Mark complete
Problem 5 A fractional output

A model gives $A(s)=5-2s-s^2$ for real $s$ . Find every input at which $A(s)=1/2$ , exactly.
1. Hint 1
  Move the target to create a standard quadratic.
2. Hint 2
  Clear the numerical fraction before applying the formula.
Answer

$s=-1\pm\frac{\sqrt{22}}2$ .

Full solution

The equation is
$s^2+2s-9/2=0.$
Multiplying by $2$ gives $2s^2+4s-9=0$ .
$\begin{aligned} \Delta&=16+72 \\ \Delta&=88 \\ s&=\frac{-4\pm\sqrt{88}}4 \\ s&=-1\pm\frac{\sqrt{22}}2 \end{aligned}$

The original rule is
$A(s)=6-(s+1)^2.$
Each candidate has squared shift $22/4=11/2$ , giving $A=1/2$ .

Answer

$s=-1\pm\frac{\sqrt{22}}2$ .

Key idea

A fraction in a target output can be cleared without changing the equation’s roots.
Problem 5: Mark complete
Problem 6 A movable horizontal line

For every real $h$ , determine how many points $y=-3x^2+6x+2$ shares with the line $y=h$ .
1. Hint 1
  Treat the line’s height as the constant in an intersection equation.
2. Hint 2
  The discriminant’s sign separates the cases.
Answer

$h<5$ : two points. $h=5$ : one point. $h>5$ : no points.

Full solution

The equation is $-3x^2+6x+(2-h)=0$ .
$\begin{aligned} \Delta&=36+12(2-h) \\ \Delta&=12(5-h) \end{aligned}$

It is positive for $h<5$ , zero at $h=5$ , and negative for $h>5$ .
These correspond to two, one, and no real intersections.
The vertex form $-3(x-1)^2+5$ confirms the dividing height.

Answer

$h<5$ : two points. $h=5$ : one point. $h>5$ : no points.

Key idea

Changing the target height changes the discriminant and the intersection count.
Problem 6: Mark complete
Problem 7 Vertex height from the discriminant

A quadratic has leading coefficient $-2$ and discriminant $24$ . Find its vertex height and determine its number of real zeros.
1. Hint 1
  The vertex height relates the discriminant to the leading coefficient.
2. Hint 2
  Root count depends on the discriminant’s sign.
Answer

Vertex height $3$ ; two distinct real zeros.

Full solution

The vertex height is
$\begin{aligned} k&=-\frac{24}{4(-2)} \\ k&=3 \end{aligned}$

The discriminant is positive, so there are two real zeros.
A downward parabola whose maximum is above zero fits that conclusion.

Answer

Vertex height $3$ ; two distinct real zeros.

Key idea

Vertex height depends on both the leading coefficient and discriminant, while root count depends on the discriminant’s sign.
Problem 7: Mark complete
Problem 8 A proposed expression

For $2x^2-3x-1=0$ , a student writes $x=3\pm\sqrt{17}/4$ . Is this a correct use of the quadratic formula? Correct the expression and explain.
1. Hint 1
  Which terms of the numerator does the $2a$ divide in the formula?
2. Hint 2
  Use the axis of symmetry to check the center of the proposed pair.
Answer

No; $x=\frac{3\pm\sqrt{17}}4$ .

Full solution

Here $a=2$ , $b=-3$ , $c=-1$ .
$\begin{aligned} \Delta&=9+8 \\ \Delta&=17 \end{aligned}$
The formula gives
$x=\frac{3\pm\sqrt{17}}4$
The student divided the radical term but left the $3$ undivided.
The correct pair is centered at $3/4$ , the axis $-b/(2a)$ , whereas the proposed pair is centered at $3$ .

Answer

No; $x=\frac{3\pm\sqrt{17}}4$ .

Key idea

The quadratic formula divides both numerator terms by the same denominator.
Problem 8: Mark complete
Problem 9 A repeated outcome

A quadratic with real coefficients has discriminant zero. A student says its graph touches the $x$ -axis at its vertex, regardless of whether it opens upward or downward. Is this correct? Explain.
1. Hint 1
  A zero discriminant makes the completed square’s constant zero.
2. Hint 2
  Write the quadratic in vertex form and examine its zeros and sign for each sign of the leading coefficient.
Answer

Yes; the repeated root is at the vertex for either opening direction.

Full solution

For $a\ne0$ and $\Delta=0$ , the constant term of vertex form, $-\Delta/(4a)$ , is $0$ , so vertex form becomes
$f(x)=a(x+\tfrac{b}{2a})^2$
It vanishes at $x=-b/(2a)$ and keeps the sign of $a$ everywhere else.
Thus it touches the axis at the vertex without crossing for either sign of $a$ .

Answer

Yes; the repeated root is at the vertex for either opening direction.

Key idea

A zero discriminant places the vertex on the axis and makes its root repeated.
Problem 9: Mark complete
Problem 10 A limiting case

For $f(x)=x^2+\sqrt{2}\,x+\frac12$ , a student computes discriminant zero and concludes that $f$ has rational roots. Is that conclusion valid? Give its root and explain.
1. Hint 1
  Check the coefficients required by the rational-root discriminant test.
2. Hint 2
  With a zero discriminant, evaluate the single value from the formula.
Answer

No; the root is $x=-\frac{\sqrt2}{2}$ , which is irrational.

Full solution

The computation $\Delta=(\sqrt2)^2-4(1)(1/2)=0$ is correct.
But the middle coefficient is irrational, so the rational-coefficient test does not apply.
The formula gives
$x=-\frac{\sqrt2}{2}$
Its square and linear term combine as $1/2-1+1/2=0$ , confirming the irrational repeated root.

Answer

No; the root is $x=-\frac{\sqrt2}{2}$ , which is irrational.

Key idea

A perfect-square discriminant certifies rational roots only when the coefficients satisfy the rationality hypothesis.
Problem 10: Mark complete

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