- Recall how the imaginary unit is used to rewrite the square root of a negative number. - Keep both choices represented by the \(\pm\) sign when you rewrite the expression.

1. Since \(i^2=-1\), \(\sqrt{-9}=3i\). 2. Therefore, the two solutions are \(x=4+3i\) and \(x=4-3i\).

\(x=4+3i\) and \(x=4-3i\)

- Think about the relationship between nonreal roots of a polynomial whose coefficients are all real. - In a complex-conjugate pair, compare the real parts and the signs of the imaginary parts.

1. Nonreal solutions of a quadratic with real coefficients occur as a complex-conjugate pair. 2. The conjugate of \(-2+5i\) is \(-2-5i\), so that is the other solution.

\(-2-5i\)

- Rewrite \(-36\) using \(i^2=-1\). - A quadratic square equation usually has two opposite square roots. - Check each proposed solution by squaring it.

1. A number whose square is \(-36\) can be written using \(i^2=-1\). 2. Since \((6i)^2=-36\) and \((-6i)^2=-36\), the two solutions are \(x=6i\) and \(x=-6i\).

\(x=\pm6i\)

- Real zeros of a quadratic correspond to x-intercepts. - Compare a crossing with a graph that just touches the x-axis. - Consider what a parabola with no x-intercepts means for its real roots.

1. Graph \(a\) crosses the x-axis twice, so its equation has two real solutions. 2. Graph \(b\) touches the x-axis once, so its equation has one repeated real solution. 3. Graph \(c\) does not meet the x-axis, so its equation has no real solutions and instead has two nonreal complex solutions.

Graph \(a\): two real solutions Graph \(b\): one repeated real solution Graph \(c\): no real solutions; two nonreal complex solutions

- What do real roots look like on a parabola's graph? - Connect the number of x-intercepts to the sign of the discriminant.

1. The parabola does not cross or touch the x-axis. 2. Therefore, \(f(x)=0\) has no real solutions, so its discriminant is negative.

Negative; the graph has no x-intercepts.

- Rearrange each equation so the square is alone. - Can a real number squared be negative? - Test a simple positive and negative number in the second equation.

1. The equation \(x^2+1=0\) requires \(x^2=-1\), which has no real solution because a real square is nonnegative. 2. The equation \(x^2-1=0\) has real solutions \(x=\pm1\). 3. Therefore, only \(x^2+1=0\) has no real solutions.

\(x^2+1=0\), because no real square equals \(-1\).

- Isolate the squared term before taking square roots. - Remember that a square equation normally has two solutions. - Express the square root of a negative number using \(i\).

1. Isolate the square: \(x^2=-5\). 2. Taking both square roots gives \(x=\pm\sqrt{-5}=\pm i\sqrt{5}\).

\(x=\pm i\sqrt{5}\)

- Look for the repeating cycle in the powers of \(i\). - Recall how to write the square root of a negative number using \(i\). - Use the quadratic formula.

1. The powers of \(i\) are \(i^1=i\), \(i^2=-1\), \(i^3=-i\), and \(i^4=1\). 2. Therefore, \(S=i-1-i+1=0\). 3. For \(x^2+4x+13=0\), the discriminant is \(4^2-4\cdot1\cdot13=-36\). 4. Since \(\sqrt{-36}=6i\), the quadratic formula gives \(x=\frac{-4\pm6i}{2}=-2\pm3i\).

a) \(S=0\) b) \(\{-2+3i, -2-3i\}\)

- Start from a factored quadratic whose factors correspond to the two given roots. - Relate the sum and product of the roots to the coefficients. - Use the fact that the two roots are complex conjugates when simplifying their sum and product.

1. Use the factored equation \((z-z_1)(z-z_2)=0\). Thus, \(b=-(z_1+z_2)\) and \(c=z_1z_2\). 2. The sum is \((2+i)+(2-i)=4\), so \(b=-4\). 3. The product is \((2+i)(2-i)=4-i^2=5\), so \(c=5\). 4. Therefore, the equation is \(z^2-4z+5=0\).

\(b=-4\), \(c=5\); the equation is \(z^2-4z+5=0\).

- The required method aims to create one squared binomial. - Keep both sides of the equation balanced when completing that square. - Once the square equals a negative real number, express its square roots using \(i\).

1. Move the constant term: \(x^2-6x=-13\). 2. Add \(9\) to both sides: \(x^2-6x+9=-4\). 3. Rewrite: \((x-3)^2=-4\). 4. Take square roots: \(x-3=\pm2i\). 5. Therefore, \(x=3\pm2i\).

\(x=3\pm2i\)

- Identify \(a\), \(b\), and \(c\) carefully before substituting. - Simplify the negative square root using \(i\). - Reduce the real and imaginary parts of the fraction before writing the final \(a+bi\) form.

1. Here \(a=2\), \(b=4\), and \(c=7\). 2. The discriminant is \(b^2-4ac=16-56=-40\). 3. The quadratic formula gives \(x=\frac{-4\pm\sqrt{-40}}{4}\). 4. Since \(\sqrt{-40}=2i\sqrt{10}\), \(x=-1\pm\frac{\sqrt{10}}{2}i\).

\(x=-1\pm\frac{\sqrt{10}}{2}i\)

- What expression is being squared? - Take both square roots of the number on the right. - Isolate \(x\) only after taking the roots.

1. Take square roots: \(x+2=\pm\sqrt{-9}=\pm3i\). 2. Subtract \(2\) from both sides: \(x=-2\pm3i\).

\(x=-2+3i\) or \(x=-2-3i\)

- Identify the three coefficients before substituting into the formula. - A negative discriminant does not mean there are no complex roots. - Simplify the radical and then the fraction.

1. The discriminant is \(D=2^2-4\cdot1\cdot6=-20\). 2. The quadratic formula gives \(x=\frac{-2\pm\sqrt{-20}}{2}\). 3. Since \(\sqrt{-20}=2i\sqrt{5}\), the roots are \(x=-1\pm i\sqrt{5}\).

\(x=-1+i\sqrt{5}\) or \(x=-1-i\sqrt{5}\)

- Determine when the imaginary part of \(z^2\) is zero. - Expand \((a+bi)^2\) and identify its two components. - A negative real square suggests a number on the imaginary axis. - Use \(i^2=-1\).

1. \(z^2=(a+bi)^2=a^2-b^2+2abi\). 2. The square is real exactly when its imaginary part is zero: \(2ab=0\). Thus, \(a=0\) or \(b=0\). 3. To solve \(z^2=-16\), the case \(b=0\) would require \(a^2=-16\), which has no real value of \(a\). 4. In the case \(a=0\), \(-b^2=-16\), so \(b=\pm4\). 5. Therefore, \(z=4i\) or \(z=-4i\).

a) \(a=0\) or \(b=0\) b) \(z\in\{4i, -4i\}\)

- Recall the conjugate-root theorem for polynomials with real coefficients. - Use the sum and product of the roots. - A negative discriminant gives nonreal roots.

1. Nonreal roots of a polynomial with real coefficients occur in conjugate pairs, so \(z_2=3+4i\). 2. By the relationships between roots and coefficients, \(p=-(z_1+z_2)=-6\). 3. Also, \(q=z_1z_2=(3-4i)(3+4i)=25\). 4. The discriminant is \(D=(-6)^2-4\cdot25=-64\). Since \(D<0\), the equation has no real solutions.

a) \(z_2=3+4i\) b) \(p=-6\) and \(q=25\) c) \(D=-64\), so there are no real solutions

- Use the quadratic formula and interpret a negative discriminant with the imaginary unit. - Compare the sum and product of the roots with the coefficients of the quadratic. - For part c, focus on how the real and imaginary parts of the two roots are related.

1. The quadratic formula gives \(z=\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm6i}{2}=2\pm3i\). 2. The sum is \((2+3i)+(2-3i)=4\). 3. The product is \((2+3i)(2-3i)=4+9=13\). 4. The two nonreal roots are complex conjugates: they have the same real part and opposite imaginary parts. Their sum and product are therefore real, consistent with the real coefficients of the quadratic.

a) \(z_1=2+3i\) and \(z_2=2-3i\) b) \(z_1+z_2=4\) and \(z_1z_2=13\) c) The roots are complex conjugates. Their imaginary parts cancel in the sum, and their product is real, which is consistent with a quadratic having real coefficients.

- For part a, compare the curve with the x-axis. - For part b, start with vertex form using an unknown vertical scale factor. - Use the graph's y-intercept to determine that factor. - For part c, isolate the squared expression before interpreting the negative square root over \(\mathbb{C}\).

1. The graph has no x-intercepts, so \(f(x)=0\) has no real solutions. 2. The graph shows vertex \((2,4)\) and y-intercept \((0,8)\). Write \(f(x)=a(x-2)^2+4\). 3. Use the y-intercept: \(8=a(0-2)^2+4\), so \(4a=4\) and \(a=1\). Thus, \(f(x)=(x-2)^2+4\). 4. Solve \((x-2)^2+4=0\): \((x-2)^2=-4\), so \(x-2=\pm2i\). 5. Therefore, \(x=2\pm2i\).

a) No real solutions b) Vertex \((2,4)\), y-intercept \((0,8)\), and \(f(x)=(x-2)^2+4\) c) \(x=2+2i\) and \(x=2-2i\)

- The sign of the discriminant determines whether a real-coefficient quadratic has real or nonreal solutions. - Write the discriminant as an expression in \(m\) before solving an inequality. - The boundary between two nonreal solutions and real solutions occurs when the discriminant is zero.

1. The discriminant is \(D=(-6)^2-4(1)(m+10)=36-4m-40=-4m-4\). 2. Two nonreal complex solutions occur when \(D<0\). Thus, \(-4m-4<0\), which gives \(m>-1\). 3. When \(m=-1\), the discriminant is \(0\), so the equation has one repeated real solution. 4. Substituting \(m=-1\) gives \(x^2-6x+9=0=(x-3)^2\), so the repeated solution is \(x=3\).

a) \(m>-1\) b) When \(m=-1\), the equation has the repeated real solution \(x=3\).

- Distinguish between “no real solutions” and “no solutions over \(\mathbb{C}\).” - Rewrite the square root of a negative number using \(i\), and simplify the positive radical separately. - After applying the quadratic formula, simplify both the real and imaginary parts of each solution.

1. A negative discriminant means there are no real solutions, but complex solutions still exist. 2. The quadratic formula gives \(x=\frac{-6\pm\sqrt{-96}}{6}\). 3. Since \(\sqrt{-96}=4\sqrt{6}i\), \(x=\frac{-6\pm4\sqrt{6}i}{6}\). 4. Simplifying gives \(x=-1\pm\frac{2\sqrt{6}}{3}i\).

The error is concluding that a negative discriminant means no solutions at all; it means no real solutions. Over \(\mathbb{C}\), the solutions are \(x=-1+\frac{2\sqrt{6}}{3}i\) and \(x=-1-\frac{2\sqrt{6}}{3}i\).

- Look first at the form in which each equation is already written. - One equation already has a squared expression isolated. - Another has leading coefficient \(1\) and can be turned into a square with little rewriting. - Save the general formula for the equation whose structure offers no simpler direct route.

1. For a), the squared expression is already isolated, so direct square-root reasoning is most natural: \(x-4=\pm3i\), giving \(x=4\pm3i\). 2. For b), completing the square is efficient: \(x^2-8x=-20\), so \((x-4)^2=-4\), giving \(x=4\pm2i\). 3. For c), the quadratic formula is the most direct general method. The discriminant is \(2^2-4\cdot3\cdot5=-56\), so \(x=\frac{-2\pm i\sqrt{56}}{6}=\frac{-1\pm i\sqrt{14}}{3}\).

a) Direct square-root reasoning; \(x=4\pm3i\) b) Completing the square; \(x=4\pm2i\) c) Quadratic formula; \(x=\frac{-1\pm i\sqrt{14}}{3}\)

- Substitute the proposed value into the entire polynomial and simplify carefully. - Keep real and imaginary parts separate while checking whether the result is zero. - After verifying one nonreal root, use the structure of polynomials with real coefficients.

1. Substitute \(3+2i\): \((3+2i)^2-6(3+2i)+13\). 2. Since \((3+2i)^2=5+12i\), the expression becomes \(5+12i-18-12i+13=0\). Thus, \(3+2i\) is a solution. 3. The polynomial has real coefficients, so nonreal roots occur in conjugate pairs. 4. Therefore, the other solution is \(3-2i\).

a) Substitution gives \(0\), so \(3+2i\) is a solution. b) The other solution is \(3-2i\) because a real-coefficient quadratic has nonreal roots in conjugate pairs.

- When you square \(2-i\), which cross terms appear? - Group real and imaginary terms separately after substituting. - When the coefficients are real, what changes if you replace \(i\) with \(-i\) throughout an expression? - What happens when you take the conjugate of \(0\)?

1. \((2-i)^2=3-4i\), and \(-4(2-i)=-8+4i\). 2. Therefore, \(f(2-i)=(3-4i)+(-8+4i)+5=0\), so Ana's claim is correct. 3. All coefficients are real, so \(f(\overline{z})=\overline{f(z)}\). In particular, \(f(2+i)=\overline{f(2-i)}=\overline{0}=0\). 4. Therefore, \(2+i\) is also a zero. This uses the conjugate property rather than the sum of roots.

a) \(f(2-i)=0\), so the claim is correct. b) \(f(2+i)=\overline{f(2-i)}=\overline{0}=0\), so \(2+i\) is also a zero.

- What second zero is forced by the real coefficients? - Write both factors before expanding. - Use the y-intercept to find the multiplier.

1. Since the coefficients are real, the other zero is \(2-i\). 2. The function has the form \(f(x)=a(x-(2+i))(x-(2-i))=a((x-2)^2+1)\). 3. At \(x=0\), \(10=a(4+1)=5a\), so \(a=2\). 4. Therefore, \(f(x)=2(x^2-4x+5)=2x^2-8x+10\).

\(f(x)=2x^2-8x+10\)

- What exactly does a negative discriminant tell you about real roots? - Rewrite the equation to make one square visible. - Include both complex square roots.

1. The discriminant is \(D=(-4)^2-4\cdot1\cdot8=-16\), so there are no real solutions, but nonreal complex solutions are possible. 2. Completing the square gives \((x-2)^2+4=0\), or \((x-2)^2=-4\). 3. Thus \(x-2=\pm2i\), so \(x=2\pm2i\).

The negative discriminant rules out real solutions, not complex ones. The solutions are \(x=2+2i\) and \(x=2-2i\).

- Substitute for the repeated expression \((z+1)^2\). - Solve the resulting quadratic equation first. - Substitute back to find \(z\). - Negative values have square roots in \(\mathbb{C}\).

1. Let \(u=(z+1)^2\). Then \(u^2-2u-8=0\). 2. Factor: \((u-4)(u+2)=0\), so \(u=4\) or \(u=-2\). 3. If \((z+1)^2=4\), then \(z+1=\pm2\), giving \(z=1\) or \(z=-3\). 4. If \((z+1)^2=-2\), then \(z+1=\pm i\sqrt{2}\), giving \(z=-1\pm i\sqrt{2}\).

\(z\in\{1, -3, -1+i\sqrt{2}, -1-i\sqrt{2}\}\)

- Express the discriminant in terms of \(k\). - Compare when that expression is positive, zero, or negative. - For nonreal roots, use the quadratic formula to identify the magnitude of each imaginary part. - Work backward from the requested imaginary parts.

1. The discriminant is \(D=(-6)^2-4(k^2+5)=16-4k^2\). 2. When \(-2<k<2\), \(D>0\) and there are two distinct real solutions. When \(k=\pm2\), \(D=0\) and there is one repeated real solution. When \(k<-2\) or \(k>2\), \(D<0\) and there are two nonreal complex solutions. 3. For \(|k|>2\), the quadratic formula gives \(x=3\pm i\sqrt{k^2-4}\). 4. Imaginary parts \(2\) and \(-2\) require \(\sqrt{k^2-4}=2\), so \(k^2=8\) and \(k=\pm2\sqrt{2}\). 5. In both cases the two solutions are \(3+2i\) and \(3-2i\).

a) Two distinct real: \(-2<k<2\); one repeated real: \(k=-2\) or \(k=2\); two nonreal complex: \(k<-2\) or \(k>2\). b) \(k=\pm2\sqrt{2}\); solutions \(3+2i\) and \(3-2i\).