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Complex Roots of Quadratics: 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 shifted square

Solve $(x-2)^2+6=0$ over the complex numbers.
1. Hint 1
  Isolate the squared expression before taking square roots.
2. Hint 2
  Both square roots of a negative real number are needed.
Answer

$x=2\pm i\sqrt6$ .

Full solution

The equation gives $(x-2)^2=-6$ , so
$x-2=\pm i\sqrt6$
Add $2$ to obtain
$x=2\pm i\sqrt6.$
Each candidate makes the square $-6$ and the original left side zero.

Answer

$x=2\pm i\sqrt6$ .

Key idea

A shifted square gives two complex roots when the isolated square is negative.
Problem 1: Mark complete
Problem 2 A compact expression

Write $2(x+3)^2+10$ as a product of linear factors over the complex numbers.
1. Hint 1
  Find the values of $x$ that make the expression zero.
2. Hint 2
  Keep the outside coefficient when writing the linear factors.
Answer

$2(x+3-i\sqrt5)(x+3+i\sqrt5)$ .

Full solution

After dividing by $2$ , a zero must satisfy $(x+3)^2=-5$ , so the roots are $-3\pm i\sqrt5$ .
Therefore the factorization is
$2(x+3-i\sqrt5)(x+3+i\sqrt5)$
Multiplying the conjugate factors gives $(x+3)^2+5$ ; the outside $2$ restores the original expression.

Answer

$2(x+3-i\sqrt5)(x+3+i\sqrt5)$ .

Key idea

Complex linear factors retain the original quadratic's leading coefficient.
Problem 2: Mark complete
Problem 3 An incomplete coefficient record

A monic quadratic with real coefficients has roots whose sum is $-10$ and whose imaginary parts are $4$ and $-4$ . Write the quadratic in standard form.
1. Hint 1
  Conjugate roots share a real part.
2. Hint 2
  Use the sum to find that real part and then compute the product of the pair.
Answer

$x^2+10x+41$ .

Full solution

The common real part is $-5$ , so the roots are $-5\pm4i$ .
Their product is
$(-5)^2+4^2=41$
A monic quadratic has negative root sum as its linear coefficient and root product as its constant, giving $x^2+10x+41$ .

Answer

$x^2+10x+41$ .

Key idea

The sum and imaginary parts of conjugate roots determine their monic quadratic.
Problem 3: Mark complete
Problem 4 Two formulas meet

Find every complex input for which $2x^2+23$ and $4x+1$ have the same value. Give the common value at each input.
1. Hint 1
  Equate the two expressions to get a quadratic equation.
2. Hint 2
  After solving the equation, use the linear expression to find each common value.
Answer

$x=1+i\sqrt{10}$ : value $5+4i\sqrt{10}$ ; $x=1-i\sqrt{10}$ : value $5-4i\sqrt{10}$ .

Full solution

Equating and simplifying gives
$2x^2-4x+22=0$
Dividing by $2$ :
$x^2-2x+11=0$
Its discriminant is $-40$ , so
$x=1\pm i\sqrt{10}.$
Substitution in $4x+1$ gives $5\pm4i\sqrt{10}$ , with matching signs.

For $x=1\pm i\sqrt{10}$ , the square is $-9\pm2i\sqrt{10}$ , so $2x^2+23$ also equals $5\pm4i\sqrt{10}$ .
This checks both the inputs and the common outputs.

Answer

$x=1+i\sqrt{10}$ : value $5+4i\sqrt{10}$ ; $x=1-i\sqrt{10}$ : value $5-4i\sqrt{10}$ .

Key idea

An equality of two expressions can reveal complex inputs even when their real graphs do not intersect.
Problem 4: Mark complete
Problem 5 A graph and its equation

The real graph in the figure belongs to a monic quadratic $f$ . Read its vertex to write $f$ in vertex form, then find its complex zeros and compare their real parts with the graph symmetry axis.

The graph of $f$ .

Text description of this figure

A grid with the horizontal x-axis from -7 to 1 at every integer and the vertical y-axis from 0 to 26, labeled at 0, 3, 6, 9, 12, 15, 18, 21 and 24. An upward-opening curve has its vertex marked at the point labeled V, which lines up with x equals negative 3 and y equals 8. No equation, coordinates, or axis of symmetry is printed.
1. Hint 1
  A monic quadratic has coefficient $1$ on its squared term in vertex form.
2. Hint 2
  A zero requires the squared displacement from the vertex to be negative.
Answer

$f(x)=(x+3)^2+8$ ; zeros $-3\pm2i\sqrt2$ ; both real parts are $-3$ , on the axis $x=-3$ .

Full solution

The vertex is $(-3,8)$ and the leading coefficient is $1$ , so
$f(x)=(x+3)^2+8$
Setting $f(x)=0$ gives $(x+3)^2=-8$ and zeros $-3\pm2i\sqrt2$ .
Both real parts are $-3$ , matching the graph axis $x=-3$ .
These complex zeros are points in the complex plane, not points on the displayed real graph.

Answer

$f(x)=(x+3)^2+8$ ; zeros $-3\pm2i\sqrt2$ ; both real parts are $-3$ , on the axis $x=-3$ .

Key idea

The shared real part of nonreal quadratic zeros is the real graph axis of symmetry.
Problem 5: Mark complete
Problem 6 One root and one value

A quadratic $g$ has real coefficients, has $-1+i\sqrt2$ as a zero, and satisfies $g(0)=9$ . Find $g$ in standard form and give its factorization into complex linear factors.
1. Hint 1
  Real coefficients supply the conjugate root.
2. Hint 2
  First build the monic product, then use $g(0)$ to set the multiplier.
Answer

$g(x)=3x^2+6x+9$ ; $g(x)=3(x+1-i\sqrt2)(x+1+i\sqrt2)$ .

Full solution

The other root is $-1-i\sqrt2$ .
Their factors multiply to
$(x+1)^2+2=x^2+2x+3$
Thus $g(x)=a(x^2+2x+3)$ with real $a\ne0$ .
Since $g(0)=3a=9$ , $a=3$ , producing $3x^2+6x+9$ and the stated linear factors.
Substituting either root makes one factor zero.

Answer

$g(x)=3x^2+6x+9$ ; $g(x)=3(x+1-i\sqrt2)(x+1+i\sqrt2)$ .

Key idea

One nonreal root fixes a real quadratic up to scale; a known nonzero function value fixes that scale.
Problem 6: Mark complete
Problem 7 A moving equation

For real $t$ , consider $x^2-2tx+t^2+4=0$ . Find its complex roots in terms of $t$ , and describe what changes and what stays fixed as $t$ varies.
1. Hint 1
  Group the first three terms into one square.
2. Hint 2
  The remaining constant controls the distance of each root from the real axis.
Answer

Roots $t\pm2i$ ; their real parts vary with $t$ , their imaginary parts stay $\pm2$ , and the graph axis is $x=t$ .

Full solution

The equation is
$(x-t)^2=-4$
Hence $x=t\pm2i$ .
Their shared real part and the real graph axis both move with $t$ , while the imaginary parts remain $2$ and $-2$ .
Squaring either displacement $x-t$ gives $-4$ , checking every real parameter value.

Answer

Roots $t\pm2i$ ; their real parts vary with $t$ , their imaginary parts stay $\pm2$ , and the graph axis is $x=t$ .

Key idea

Changing the center of a completed square changes the shared real part of its nonreal roots.
Problem 7: Mark complete
Problem 8 A proposed factorization

A student writes $x^2+2x+4=(x+1-i)(x+1+i)$ . Decide whether this is correct; if not, give a correct factorization into complex linear factors.
1. Hint 1
  Expand the proposed product and compare its constant term.
2. Hint 2
  Write the original polynomial as $(x+1)^2$ plus a constant.
Answer

Incorrect; $(x+1-i\sqrt3)(x+1+i\sqrt3)$ .

Full solution

The proposed factors multiply to $(x+1)^2+1=x^2+2x+2$ , whose constant is wrong.
The original expression is
$(x+1)^2+3$
Since $(i\sqrt3)^2=-3$ , this factors as $(x+1-i\sqrt3)(x+1+i\sqrt3)$ .
Expanding restores the constant $4$ .

Answer

Incorrect; $(x+1-i\sqrt3)(x+1+i\sqrt3)$ .

Key idea

The imaginary part of a root comes from the leftover constant after completing the square.
Problem 8: Mark complete
Problem 9 A coefficient condition

A student claims that if $z$ solves $az^2+bz+c=0$ with real $a,b,c$ and $a\ne0$ , then $\overline z$ solves the same equation. Prove the claim using conjugation, and explain which step depends on the coefficients being real.
1. Hint 1
  Conjugation passes through sums and products.
2. Hint 2
  Conjugate the whole equation, including the coefficients and zero.
Answer

The claim is true; the conjugated equation is $a\overline z^{\,2}+b\overline z+c=0$ ; the step replacing $\overline a,\overline b,\overline c$ by $a,b,c$ is the one that needs real coefficients.

Full solution

Conjugate both sides of the given equality.
$\overline{az^2+bz+c}=0$
Passing the conjugate through the operations gives
$\overline a\,\overline z^{\,2}+\overline b\,\overline z+\overline c=0$
Real coefficients equal their own conjugates, so this becomes
$a\overline z^{\,2}+b\overline z+c=0.$
That coefficient replacement is the step requiring real coefficients.

Answer

The claim is true; the conjugated equation is $a\overline z^{\,2}+b\overline z+c=0$ ; the step replacing $\overline a,\overline b,\overline c$ by $a,b,c$ is the one that needs real coefficients.

Key idea

Real coefficients remain unchanged by conjugation, forcing nonreal roots to occur in conjugate pairs.
Problem 9: Mark complete
Problem 10 A count in two systems

A report says the equation $(x-4)^2+2=0$ has no roots, but a second report says it has two. Resolve the disagreement, giving the roots and explaining what a negative discriminant says once complex numbers are allowed.
1. Hint 1
  Each report needs to state its number system.
2. Hint 2
  A negative square is impossible for real numbers but possible for complex numbers.
Answer

No real roots; two complex roots $4\pm i\sqrt2$ . A negative discriminant places the roots off the real line.

Full solution

The equation gives $(x-4)^2=-2$ , which has no real solution.
Over the complex numbers it gives
$x=4\pm i\sqrt2$
For either root, $x-4=\pm i\sqrt2$ , so
$(x-4)^2=-2.$
The expanded quadratic has discriminant $-8$ , which specifies a distinct nonreal conjugate pair; it no longer means that roots are missing from the complex system.

Answer

No real roots; two complex roots $4\pm i\sqrt2$ . A negative discriminant places the roots off the real line.

Key idea

The number system determines whether a quadratic refusal is about all roots or just real roots.
Problem 10: Mark complete

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