This algebra 2 quadratics worksheet answer key solves all 24 problems of the quadratic equations with complex solutions worksheet step by step.
Complete worked solutions for the worksheet. Each answer shows the discriminant or the key step, the simplified solutions in bold, and a graph or complex-plane picture when it helps. Check that complex solutions are written in the form a + bi.
Algebra 2 Quadratics Worksheet Answer Key: problems 1–5
Problem 1: Square root method
Take the square root of both sides: \(x=\pm\sqrt{-49}=\pm 7i\).
\(x=\pm 7i\)
Problem 2: Isolate the square
\(3x^{2}=-27\), so \(x^{2}=-9\) and \(x=\pm 3i\).
\(x=\pm 3i\)
Problem 3: Square of a binomial
\(x-2=\pm 5i\), so \(x=2\pm 5i\).
\(x=2\pm 5i\)
Problem 4: Complete the isolation
\(2(x+1)^{2}=-18\), \((x+1)^{2}=-9\), \(x+1=\pm 3i\).
\(x=-1\pm 3i\)
Problem 5: Compute discriminants
a) \(D=36-20=16\). b) \(D=36-36=0\). c) \(D=36-52=-16\).
a) 16, two real solutions b) 0, one repeated real solution c) \(-16\), two complex solutions
Algebra 2 Quadratics Worksheet Answer Key: problems 6–10
Problem 6: Match graphs to discriminants
A graph that crosses the x-axis twice means two real solutions. One that touches it once means one repeated solution. One that stays above or below it means no real solutions.
I: \(D > 0\), 2 real solutions. II: \(D=0\), 1 real solution. III: \(D < 0\), 0 real solutions (two complex).
Problem 7: Graph and solve
The vertex is \((-2,1)\) and the parabola opens up, so it never reaches the x-axis: no real solutions.
\(D=16-20=-4\), so \(x=\frac{-4\pm 2i}{2}=-2\pm i\).
No real solutions; \(x=-2\pm i\)
Problem 8: Quadratic formula
\(D=4-40=-36\). \(x=\frac{2\pm 6i}{2}=1\pm 3i\).
\(x=1\pm 3i\)
Problem 9: Complete the square
\(x^{2}+6x=-13\). Add \(9\): \((x+3)^{2}=-4\). Then \(x+3=\pm 2i\).
\(x=-3\pm 2i\)
Problem 10: Larger coefficients
\(D=64-100=-36\). \(x=\frac{8\pm 6i}{2}=4\pm 3i\).
\(x=4\pm 3i\)
Algebra 2 Quadratics Worksheet Answer Key: problems 11–15
Problem 11: Leading coefficient not 1
\(D=16-40=-24\) and \(\sqrt{-24}=2i\sqrt{6}\). \(x=\frac{4\pm 2i\sqrt{6}}{4}=1\pm\frac{\sqrt{6}}{2}i\).
\(x=1\pm\frac{\sqrt{6}}{2}i\)
Problem 12: Fractions in the answer
\(D=4-12=-8\) and \(\sqrt{-8}=2i\sqrt{2}\). \(x=\frac{-2\pm 2i\sqrt{2}}{6}=\frac{-1\pm i\sqrt{2}}{3}\).
\(x=\frac{-1\pm i\sqrt{2}}{3}\)
Problem 13: Rearrange first
Write it as \(x^{2}-2x+17=0\). \(D=4-68=-64\). \(x=\frac{2\pm 8i}{2}=1\pm 4i\).
\(x=1\pm 4i\)
Problem 14: Expand then solve
\(x^{2}+6x+10=0\). \(D=36-40=-4\). \(x=\frac{-6\pm 2i}{2}=-3\pm i\).
\(x=-3\pm i\)
Problem 15: Find a constant
\(D=36-4k\).
a) \(36-4k=0\) gives \(k=9\). b) \(36-4k < 0\) gives \(k > 9\).
a) \(k=9\) b) \(k > 9\)
Algebra 2 Quadratics Worksheet Answer Key: problems 16–20
Problem 16: Verify a complex solution
\((1-2i)^{2}=1-4i+4i^{2}=-3-4i\) and \(-2(1-2i)=-2+4i\).
\((-3-4i)+(-2+4i)+5=0\).
The left side is 0, so \(1-2i\) is a solution.
Problem 17: Build an equation from its roots
The roots are \(3+i\) and \(3-i\). Their sum is \(6\) and their product is \(9-i^{2}=10\). For \(x^{2}+bx+c\), the sum is \(-b\) and the product is \(c\).
\(b=-6\), \(c=10\); the equation is \(x^{2}-6x+10=0\).
Problem 18: Conjugate roots
Complex roots of a real quadratic are conjugates, so the other root is \(2+5i\). Sum: \(4=-b\), so \(b=-4\). Product: \(4+25=29\), so \(c=29\).
Other root \(2+5i\); \(b=-4\), \(c=29\)
Problem 19: Does the ball reach 100 feet?
\(-16t^{2}+64t=100\) becomes \(4t^{2}-16t+25=0\) after moving terms and dividing by \(-4\).
\(D=256-400=-144\), so \(t=\frac{16\pm 12i}{8}=2\pm 1.5i\).
The solutions are not real, so the ball never reaches 100 ft. Its maximum height is \(h(2)=64\) ft.
Problem 20: Revenue at the school store
\(-5p^{2}+100p=600\) gives \(p^{2}-20p+120=0\). \(D=400-480=-80\).
\(p=\frac{20\pm 4i\sqrt{5}}{2}=10\pm 2i\sqrt{5}\).
The solutions are not real, so no price gives \(\$600\). The maximum revenue is \(\$500\) at \(p=\$10\).
Algebra 2 Quadratics Worksheet Answer Key: problems 21–24
Problem 21: Find the error
He treated \(\sqrt{-4}\) as 2, but \(\sqrt{-4}=2i\). Check \(x=4\): \(16-24+10=2\neq 0\).
Correct: \(x=\frac{6\pm 2i}{2}=3\pm i\).
\(x=3\pm i\)
Problem 22: Garden with perimeter 20 feet
Length is \(10-w\), so \(w(10-w)=30\) gives \(w^{2}-10w+30=0\).
\(D=100-120=-20\), so \(w=\frac{10\pm 2i\sqrt{5}}{2}=5\pm i\sqrt{5}\).
The width would not be real, so the garden is impossible. The greatest possible area is \(5\cdot 5=25\) sq ft.
Problem 23: Integer values of k
\(D=k^{2}-64\). Non-real solutions need \(k^{2}-64 < 0\), so \(k^{2} < 64\) and \(-8 < k < 8\).
The integers from \(-7\) to \(7\) work, which is 15 values.
\(-7\le k\le 7\), 15 integers
Problem 24: Sum and product of solutions
\(D=64-88=-24\) and \(\sqrt{-24}=2i\sqrt{6}\). \(x=\frac{-8\pm 2i\sqrt{6}}{4}=-2\pm\frac{\sqrt{6}}{2}i\).
Sum: \(-2+(-2)=-4=-\frac{8}{2}\). Product: \((-2)^{2}-(\frac{\sqrt{6}}{2}i)^{2}=4+\frac{6}{4}=\frac{11}{2}\).
\(x=-2\pm\frac{\sqrt{6}}{2}i\); sum \(-4\), product \(\frac{11}{2}\)
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