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.
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
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
\(D=64-100=-36\). \(x=\frac{8\pm 6i}{2}=4\pm 3i\).
\(x=4\pm 3i\)
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) \(k > 9\)
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 \(p=\$10\).
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 \(\frac{11}{2}\)
See every chapter on the Algebra 2 math page.