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The Quadratic Formula: 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 radical to simplify

Solve $3x^2-4x-2=0$ with the quadratic formula, and give both roots in simplest form.
1. Hint 1
  The formula needs the three coefficients read with their signs, so name $a$ , $b$ and $c$ before substituting.
2. Hint 2
  Simplify the radical first, then divide the whole numerator by $2a$ and reduce what is left.
Answer

$x=\dfrac{2+\sqrt{10}}3$ and $x=\dfrac{2-\sqrt{10}}3$ , written together as $x=\dfrac{2\pm\sqrt{10}}3$ .

Full solution

Read the coefficients with their signs: $a=3$ , $b=-4$ and $c=-2$ .
The discriminant is
$D=16+24$

So $D=40$ , positive but not a perfect square.
Substituting, with $-b=4$ and $2a=6$ , gives
$x=\frac{4\pm\sqrt{40}}{6}$

Since $40=4\times10$ , the radical becomes $2\sqrt{10}$ .
$x=\frac{4\pm2\sqrt{10}}{6}$

Both numerator terms and the denominator carry a factor $2$ , so divide it out.
$x=\frac{2\pm\sqrt{10}}{3}$

Each root gives $3x-2=\pm\sqrt{10}$ , so $9x^2-12x-6=0$ , which is three times the original equation.

Answer

$x=\dfrac{2+\sqrt{10}}3$ and $x=\dfrac{2-\sqrt{10}}3$ , written together as $x=\dfrac{2\pm\sqrt{10}}3$ .

Key idea

Simplify the radical first, then reduce only by a factor shared by both numerator terms and $2a$ .
Problem 1: Mark complete
Problem 2 Discriminant, then roots

Solve $5x^2+4x+2=0$ with the quadratic formula, reporting the discriminant and both roots in simplest form.
1. Hint 1
  The sign of $b^2-4ac$ settles what kind of numbers the roots are before any simplifying begins.
2. Hint 2
  A negative quantity under the radical is rewritten with $i$ , and the whole numerator is then divided by $2a$ .
Answer

$D=-24$ ; $x=\dfrac{-2+i\sqrt6}5$ and $x=\dfrac{-2-i\sqrt6}5$ , written together as $x=\dfrac{-2\pm i\sqrt6}5$ , or equivalently $x=-\dfrac25\pm\dfrac{\sqrt6}5 i$ .

Full solution

Read the coefficients: $a=5$ , $b=4$ and $c=2$ .
The discriminant is
$D=16-40$

So $D=-24$ .
A negative discriminant means the roots are nonreal, and substituting gives
$x=\frac{-4\pm\sqrt{-24}}{10}$

Rewrite the negative under the radical with $i$ , since $\sqrt{-24}=2i\sqrt6$ .
$x=\frac{-4\pm2i\sqrt6}{10}$

Every part of the numerator and the denominator carries a factor $2$ , so divide it out.
$x=\frac{-2\pm i\sqrt6}{5}$

The two roots share the real part $-2/5$ and differ only in the sign of the imaginary part, the conjugate pair a negative discriminant produces.

Answer

$D=-24$ ; $x=\dfrac{-2+i\sqrt6}5$ and $x=\dfrac{-2-i\sqrt6}5$ , written together as $x=\dfrac{-2\pm i\sqrt6}5$ , or equivalently $x=-\dfrac25\pm\dfrac{\sqrt6}5 i$ .

Key idea

A negative discriminant sends the formula through $i$ and returns a conjugate pair.
Problem 2: Mark complete
Problem 3 Roots of $3x^2+4=8x$

Use the quadratic formula to solve $3x^2+4=8x$ , giving both roots in simplest form.
1. Hint 1
  The three coefficients can be read only once every term sits on one side with zero on the other.
2. Hint 2
  Work out the discriminant before writing any radical, and check whether it is a perfect square.
Answer

$x=2$ and $x=\dfrac23$ .

Full solution

Move every term to one side so that the other side is zero.
$3x^2-8x+4=0$

Now $a=3$ , $b=-8$ and $c=4$ , so the discriminant is
$D=64-48$

So $D=16$ , which is a perfect square, and
$\sqrt{16}=4.$
With $-b=8$ and $2a=6$ , substituting gives
$x=\frac{8\pm4}{6}$

The plus sign gives $12/6=2$ , and the minus sign gives $4/6=2/3$ .
No radical survives, so nothing is left to simplify.

A perfect-square discriminant makes the roots rational, and indeed $3x^2-8x+4=(3x-2)(x-2)$ , which is zero at exactly those two values.

Answer

$x=2$ and $x=\dfrac23$ .

Key idea

Coefficients are readable only once an equation is set equal to zero, so standard form comes first.
Problem 3: Mark complete
Problem 4 Equal values

Find every real $x$ at which the expressions $2x^2-3x+1$ and $x^2+2x+4$ take the same value, and classify the roots of the quadratic equation you solve to find them.
1. Hint 1
  Two expressions take the same value exactly where their difference is zero.
2. Hint 2
  Identify the three coefficients, evaluate the discriminant, and simplify the formula.
Answer

$x=\dfrac{5\pm\sqrt{37}}2$ ; two distinct real roots.

Full solution

Subtract the second expression from the first.
$x^2-5x-3$

The two values agree exactly where this difference is zero.
$x^2-5x-3=0$

The coefficients are $a=1$ , $b=-5$ , and $c=-3$ .
$D=25+12$

Thus $D=37>0$ , which gives two distinct real roots.
The formula gives
$x=\frac{5\pm\sqrt{37}}2$

Each value satisfies $(x-5/2)^2=37/4$ , confirming the difference of the two expressions is zero.

Answer

$x=\dfrac{5\pm\sqrt{37}}2$ ; two distinct real roots.

Key idea

Where two quadratic expressions differ in their squared terms, setting them equal leaves a quadratic whose discriminant and formula settle where they agree.
Problem 4: Mark complete
Problem 5 Checking a student's line

Solving $2x^2+6x-1=0$ , a student writes $x=-6\pm\dfrac{2\sqrt{11}}4$ . Identify the error in that line, and give the correct roots in simplest form.
1. Hint 1
  Compare the student's line with $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ piece by piece: the $-b$ term, the radical, and the denominator.
2. Hint 2
  The fraction bar in the formula runs under the whole numerator, so ask what $2a$ has to divide.
3. Hint 3
  Once the correct fraction is written, reduce it by the factor its numerator shares with $4$ .
Answer

The student divided only the radical by $2a=4$ ; the whole numerator $-6\pm2\sqrt{11}$ goes over $4$ . The correct roots are $x=\dfrac{-3\pm\sqrt{11}}2$ .

Full solution

Read the coefficients: $a=2$ , $b=6$ and $c=-1$ .
The discriminant is
$D=36+8$

So $D=44$ , and $\sqrt{44}=2\sqrt{11}$ ; the student found both of those correctly.
The formula puts the entire numerator over $2a=4$ .
$x=\frac{-6\pm2\sqrt{11}}{4}$

The student divided only the radical by $4$ and left the $-6$ undivided, which moves both roots.
Dividing the numerator and the denominator by $2$ finishes the correct line.
$x=\frac{-3\pm\sqrt{11}}{2}$

Each correct root gives $2x+3=\pm\sqrt{11}$ , so $4x^2+12x-2=0$ , which is twice the original equation.
The student's two values are about $-4.34$ and $-7.66$ , and neither satisfies it.

Answer

The student divided only the radical by $2a=4$ ; the whole numerator $-6\pm2\sqrt{11}$ goes over $4$ . The correct roots are $x=\dfrac{-3\pm\sqrt{11}}2$ .

Key idea

The bar in the formula divides the whole numerator, so $-b$ and the radical are divided together.
Problem 5: Mark complete
Problem 6 A negative leading coefficient

Solve $-2x^2+6x+7=0$ with the quadratic formula exactly as it stands, taking $a=-2$ so that $2a$ is negative. Then multiply every term by $-1$ and solve $2x^2-6x-7=0$ the same way.

Give the solution set each substitution produces, and explain why the negative denominator does not change which two numbers come out.
1. Hint 1
  The formula never asked $a$ to be positive, so substitute the coefficients as they stand and carry the sign of $2a$ with them.
2. Hint 2
  Simplify the radical before dividing, then divide both numerator terms by the negative denominator.
3. Hint 3
  Ask which branch of the $\pm$ becomes the larger root once the denominator is negative.
Answer

Both substitutions give the same solution set, $x=\dfrac{3+\sqrt{23}}2$ and $x=\dfrac{3-\sqrt{23}}2$ . The negative denominator only exchanges which branch of the $\pm$ produces which root.

Full solution

Read the first equation as it stands: $a=-2$ , $b=6$ and $c=7$ , so $-4ac=56$ and the discriminant is
$D=36+56$

So $D=92$ .
Since $92=4\times23$ , the radical simplifies to
$\sqrt{92}=2\sqrt{23}$

Here $-b=-6$ and $2a=-4$ , so the whole numerator sits over $-4$ .
$x=\frac{-6\pm2\sqrt{23}}{-4}$

Divide both numerator terms by $-4$ .
$x=\frac{3\mp\sqrt{23}}{2}$

The upper sign now belongs to $3-\sqrt{23}$ , because dividing by a negative number reverses which branch of the $\pm$ is larger.
The two roots are $(3+\sqrt{23})/2$ and $(3-\sqrt{23})/2$ .

Multiplying every term of that equation by $-1$ gives the second equation, with $a=2$ , $b=-6$ and $c=-7$ .
Its discriminant is $36+56=92$ again, and substituting gives
$x=\frac{6\pm2\sqrt{23}}{4}$

Dividing by the positive $4$ returns the same two numbers, this time with the upper sign on the larger root.
$x=\frac{3\pm\sqrt{23}}{2}$

Multiplying an equation by a nonzero number cannot change which values make it true, so the two substitutions had to land on the same pair; only the bookkeeping of the $\pm$ differed.

Answer

Both substitutions give the same solution set, $x=\dfrac{3+\sqrt{23}}2$ and $x=\dfrac{3-\sqrt{23}}2$ . The negative denominator only exchanges which branch of the $\pm$ produces which root.

Key idea

Multiplying every coefficient by a nonzero number leaves the roots alone, so a quadratic with a negative leading coefficient can be solved as it stands or flipped first.
Problem 6: Mark complete
Problem 7 A specified separation

The equation $2x^2-8x+c=0$ has two real roots whose difference, larger minus smaller, is $3$ . Find $c$ and the two roots.

Then increase the value of $c$ by $9/2$ and use the discriminant to predict the type of roots of the modified equation.
1. Hint 1
  Two real roots sit the same distance either side of $-b/(2a)$ , so the gap between them comes from the discriminant and $a$ alone.
2. Hint 2
  Write the two roots with the formula while leaving $c$ unknown.
3. Hint 3
  Subtract the smaller formula value from the larger to relate the discriminant to the given separation.
4. Hint 4
  Recalculate the discriminant after changing $c$ , and use its sign to classify the new roots.
Answer

$c=\dfrac72$ ; the original roots are $\dfrac12$ and $\dfrac72$ . The modified equation has one repeated real root.

Full solution

The discriminant is
$D=64-8c$

The formula gives
$x=(8\pm\sqrt D)/4.$
Subtracting the smaller root from the larger gives
$\frac{2\sqrt D}{4}=3$

Hence $\sqrt D=6$ and $D=36$ .
Solving $64-8c=36$ gives $c=7/2$ .
The roots are
$(8\pm6)/4$

They are $1/2$ and $7/2$ , whose difference is $3$ .
Both satisfy
$2x^2-8x+7/2=0.$

The increased constant is $7/2+9/2=8$ .
Its discriminant is
$D=64-8(8)$

This is zero, so the modified equation has one repeated real root.

Answer

$c=\dfrac72$ ; the original roots are $\dfrac12$ and $\dfrac72$ . The modified equation has one repeated real root.

Key idea

The separation of two real roots can be read from the square-root terms in their formula expressions.
Problem 7: Mark complete
Problem 8 Opposite signs

For real coefficients with $a\ne0$ , a student claims that $ax^2+bx+c=0$ has two distinct real roots whenever $a$ and $c$ have opposite signs. Is the claim correct for every real $b$ ? Explain.
1. Hint 1
  The root type is decided by the discriminant alone, so ask what a sign condition on $a$ and $c$ forces the term $-4ac$ to do.
2. Hint 2
  Compare the discriminant with $b^2$ , including the possibility $b=0$ .
Answer

Yes, the claim is correct for every real $b$ .

Full solution

Opposite signs give $ac<0$ , so $-4ac>0$ .
The discriminant is
$D=b^2-4ac$

The quantity $b^2$ is zero or positive, and the added quantity $-4ac$ is positive.
Thus $D>0$ , even when $b=0$ .
A positive discriminant gives two distinct real roots.

Answer

Yes, the claim is correct for every real $b$ .

Key idea

Opposite signs on the leading and constant coefficients force a positive discriminant.
Problem 8: Mark complete
Problem 9 Matching discriminants

A student claims that two monic quadratic equations with the same discriminant must have the same roots. Decide whether the claim is correct, justify your decision with two monic quadratic equations and the roots of each, and say what the discriminant of a monic quadratic determines about its roots.
1. Hint 1
  Equal discriminants leave the rest of the formula free, so look at what the middle coefficient can still do.
2. Hint 2
  For a monic quadratic the roots are $-b/2\pm\sqrt D/2$ , so ask which part of that the discriminant fixes.
3. Hint 3
  Keep the leading coefficient $1$ and the same discriminant, and change only the sign of the middle coefficient.
Answer

False. For example $x^2-2x-2=0$ and $x^2+2x-2=0$ both have discriminant $12$ , with roots $1\pm\sqrt3$ and $-1\pm\sqrt3$ . A discriminant fixes the root type and the size of the $\pm$ term, not the roots.

Full solution

Take two equations with leading coefficient $1$ whose discriminants agree, for instance $x^2-2x-2=0$ and
$x^2+2x-2=0.$
For each of them
$D=4+8$

So each discriminant is $12$ , and
$\sqrt{12}=2\sqrt3.$
The first equation has $-b=2$ , so the formula gives
$x=\frac{2\pm2\sqrt3}{2}$

Both numerator terms carry the factor $2$ , so that pair is $1\pm\sqrt3$ .
The second equation has $-b=-2$ , giving
$x=\frac{-2\pm2\sqrt3}{2}$

That pair reduces to $-1\pm\sqrt3$ .
The two pairs carry the same $\pm\sqrt3$ but start from $-b=2$ and $-b=-2$ , so they are shifted apart by $2$ and cannot coincide.

The claim is therefore false: a discriminant fixes the kind of roots and the size of the square-root term, while $b$ decides where that term is measured from.

Answer

False. For example $x^2-2x-2=0$ and $x^2+2x-2=0$ both have discriminant $12$ , with roots $1\pm\sqrt3$ and $-1\pm\sqrt3$ . A discriminant fixes the root type and the size of the $\pm$ term, not the roots.

Key idea

A discriminant gives root type and part of the formula, but the other coefficients also determine the roots.
Problem 9: Mark complete
Problem 10 A translated unknown

Let $a,b,c$ be real with $a\ne0$ . In $ax^2+bx+c=0$ , replace $x$ by $y+s$ . Choose $s$ so the expanded equation has no term in $y$ to the first power, then derive the quadratic formula by solving the resulting equation for $y$ and returning to $x$ .
1. Hint 1
  Expand after the replacement and collect the coefficient of $y$ .
2. Hint 2
  Set that coefficient to zero; substitute the resulting $s$ into the constant part.
3. Hint 3
  Isolate $y^2$ , keep both square roots, and use $x=y+s$ .
Answer

$s=-\dfrac{b}{2a}$ ; $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ .

Full solution

After replacing $x$ by $y+s$ , expand the two terms that carry the unknown.
$\begin{aligned} a(y+s)^2 &= ay^2+2asy+as^2 \\ b(y+s) &= by+bs \end{aligned}$

Adding these two results and the constant $c$ gives an equation in $y$ whose coefficient of $y$ is $2as+b$ and whose constant term is $as^2+bs+c$ .

The linear term vanishes when $2as+b=0$ .
Since $a\ne0$ , this gives
$s=-\frac b{2a}$

For that shift, the constant becomes $c-b^2/(4a)$ .
The equation therefore gives
$ay^2=\frac{b^2}{4a}-c$

Divide by the nonzero $a$ .
$y^2=\frac{b^2-4ac}{4a^2}$

Taking both square roots gives
$y=\pm\sqrt{b^2-4ac}/(2a).$
The positive square root of $4a^2$ is $2|a|$ , but allowing both signs gives the same pair with $2a$ .
For a negative discriminant the numerator is interpreted using $i$ .

Finally, return to $x=y+s$ .
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The substitution removed the linear term and made the remaining equation a square equation, which establishes the formula.

Answer

$s=-\dfrac{b}{2a}$ ; $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ .

Key idea

Shifting the unknown to remove the linear term turns a general quadratic into a square equation.
Problem 10: Mark complete

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