Complex Solutions of Quadratic Equations: CCSS.Math.Content.HSN-CN.C.7 - Common Core: High School - Number and Quantity

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Question

Solve for .

Answer

This question is testing one's ability to solve quadratic equations using the quadratic formula. Furthermore, it builds one's ability to recognize and understand properties of parabolas, dealing with quadratics, and identifying roots.

For the purpose of Common Core Standards, "solve quadratic equations with real coefficients that have complex solutions", falls within the Cluster C of "use complex numbers in polynomial identities and equations" (CCSS.MATH.CONTENT.HSF.CN.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: State the general quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

where the quadratic function is in the form,

.

Step 2: Identify the coefficients.

$\a=4 \b=4 \c=2$

Step 3: Substitute coefficients into the quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-4\pm\sqrt{4^2-4(4)(2)}}{2(4)}$

$x=\frac{-4\pm\sqrt{16-32}}{8}$

$x=\frac{-4\pm\sqrt{-16}}{8}$

Recall that a negative sign under the radical represents an imaginary number.

$x=\frac{-4\pm4i}{8}$

$x=\frac{-4}{8}\pm\frac{4i}{8}$

$x=\frac{-1}{2}\pm\frac{i}{2}$

This solution shows that the roots of this particular quadratic are imaginary.

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Question

Solve.

$3x^{2}-2x+7$

Answer

$\frac{2\pm \sqrt{\left ( -2 \right )^{2}-4\cdot 3\cdot 7}}{2\cdot 3}$

$\frac{2\pm \sqrt{4-84}}{6}$

$\frac{2\pm \sqrt{-80}}{6}$

$\frac{2\pm \sqrt{16\cdot 5\cdot -1}}{6}$

$\frac{2\pm 4i\sqrt{5}}{6}$

$\frac{1\pm 2i\sqrt{5}}{3}$

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Question

Solve.

$4x^{2}+2x+3$

Answer

$x=\frac{-4\pm \sqrt{\left ( 2 \right )^{2}-4\cdot 4\cdot 3}}{2\cdot 4}$

$x=\frac{-4\pm \sqrt{4-48}}{8}$

$x=\frac{-4\pm \sqrt{-44}}{8}$

$x=\frac{-4\pm \sqrt{4\cdot 11\cdot -1}}{8}$

$x=\frac{-4\pm 2i\sqrt{11}}{8}$

$x=\frac{-2\pm i\sqrt{11}}{4}$

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Question

Solve.

$x^{2}-2x+5$

Answer

$x=\frac{-(-2)\pm \sqrt{(-2)^{2}-4\cdot 1\cdot 5}}{2\cdot 1}$

$x=\frac{2\pm \sqrt{4-20}}{2}$

$x=\frac{2\pm \sqrt{-16}}{2}$

$x=\frac{2\pm 4i}{2}$

$x=1\pm 2i$

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Question

Solve.

$a^{2}-6a+13$

Answer

$a=\frac{-(-6)\pm \sqrt{\left ( -6 \right )^{2}-4\cdot 1\cdot 13}}{2\cdot 1}$

$a=\frac{6\pm \sqrt{36-52}}{2}$

$a=\frac{6\pm \sqrt{-16}}{2}$

$a=\frac{6\pm \sqrt{16\cdot -1}}{2}$

$a=\frac{6\pm 4i}{2}$

$a=3\pm 2i$

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Question

Solve.

$3b^{2}-4b+5$

Answer

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Question

Solve.

$g^{2}-18g+84$

Answer

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Question

Solve.

$2m^{2}-8m+25$

Answer

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Question

Solve.

$4a^{2}+25=0$

Answer

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Question

Solve.

$8z^{2}+72=0$

Answer

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Question

Solve.

$2x^{2}+8=0$

Answer

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Question

Solve.

$10x^{2}+10=0$

Answer

$10\left ( x^{2}+1 \right )=0$

$x^{2}+1=0$

$x^{2}=-1$

$x=\pm \sqrt{-1}$

$x=\pm i$

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Question

Solve for .

Answer

This question is testing one's ability to solve quadratic equations using the quadratic formula. Furthermore, it builds one's ability to recognize and understand properties of parabolas, dealing with quadratics, and identifying roots.

For the purpose of Common Core Standards, "solve quadratic equations with real coefficients that have complex solutions", falls within the Cluster C of "use complex numbers in polynomial identities and equations" (CCSS.MATH.CONTENT.HSF.CN.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: State the general quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

where the quadratic function is in the form,

.

Step 2: Identify the coefficients.

$\a=4 \b=4 \c=2$

Step 3: Substitute coefficients into the quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-4\pm\sqrt{4^2-4(4)(2)}}{2(4)}$

$x=\frac{-4\pm\sqrt{16-32}}{8}$

$x=\frac{-4\pm\sqrt{-16}}{8}$

Recall that a negative sign under the radical represents an imaginary number.

$x=\frac{-4\pm4i}{8}$

$x=\frac{-4}{8}\pm\frac{4i}{8}$

$x=\frac{-1}{2}\pm\frac{i}{2}$

This solution shows that the roots of this particular quadratic are imaginary.

Compare your answer with the correct one above

Question

Solve.

$3x^{2}-2x+7$

Answer

$\frac{2\pm \sqrt{\left ( -2 \right )^{2}-4\cdot 3\cdot 7}}{2\cdot 3}$

$\frac{2\pm \sqrt{4-84}}{6}$

$\frac{2\pm \sqrt{-80}}{6}$

$\frac{2\pm \sqrt{16\cdot 5\cdot -1}}{6}$

$\frac{2\pm 4i\sqrt{5}}{6}$

$\frac{1\pm 2i\sqrt{5}}{3}$

Compare your answer with the correct one above

Question

Solve.

$4x^{2}+2x+3$

Answer

$x=\frac{-4\pm \sqrt{\left ( 2 \right )^{2}-4\cdot 4\cdot 3}}{2\cdot 4}$

$x=\frac{-4\pm \sqrt{4-48}}{8}$

$x=\frac{-4\pm \sqrt{-44}}{8}$

$x=\frac{-4\pm \sqrt{4\cdot 11\cdot -1}}{8}$

$x=\frac{-4\pm 2i\sqrt{11}}{8}$

$x=\frac{-2\pm i\sqrt{11}}{4}$

Compare your answer with the correct one above

Question

Solve.

$x^{2}-2x+5$

Answer

$x=\frac{-(-2)\pm \sqrt{(-2)^{2}-4\cdot 1\cdot 5}}{2\cdot 1}$

$x=\frac{2\pm \sqrt{4-20}}{2}$

$x=\frac{2\pm \sqrt{-16}}{2}$

$x=\frac{2\pm 4i}{2}$

$x=1\pm 2i$

Compare your answer with the correct one above

Question

Solve.

$a^{2}-6a+13$

Answer

$a=\frac{-(-6)\pm \sqrt{\left ( -6 \right )^{2}-4\cdot 1\cdot 13}}{2\cdot 1}$

$a=\frac{6\pm \sqrt{36-52}}{2}$

$a=\frac{6\pm \sqrt{-16}}{2}$

$a=\frac{6\pm \sqrt{16\cdot -1}}{2}$

$a=\frac{6\pm 4i}{2}$

$a=3\pm 2i$

Compare your answer with the correct one above

Question

Solve.

$3b^{2}-4b+5$

Answer

Compare your answer with the correct one above

Question

Solve.

$g^{2}-18g+84$

Answer

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Question

Solve.

$2m^{2}-8m+25$

Answer

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