Find the complex roots of a quadratic using the discriminant Trigonometry

Consider the following quadratic equation:

ax^2+bx+c=0

The solutions are found using the quadratic formula:

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

Let's find the roots of:

P\left(x\right)=x^2+6x+10

We use:

• a=1
• b=6
• c=10

This gives:

x=\dfrac{-6\pm\sqrt{6^2-4\left(1\right)\left(10\right)}}{2\cdot1}

x=\dfrac{-6\pm\sqrt{-4}}{2}

x=\dfrac{-6\pm 2i}{2}

x=-3\pm i

Consider the following quadratic equation:

ax^2+bx+c=0

The solutions are found using the quadratic formula:

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

Let's find the roots of:

P\left(x\right)=x^2+2x+4

We use:

• a=1
• b=2
• c=4

This gives:

x=\dfrac{-2\pm\sqrt{2^2-4\left(1\right)\left(4\right)}}{2\cdot1}

x=\dfrac{-2\pm\sqrt{-12}}{2}

x=\dfrac{-2\pm 2i\sqrt{3}}{2}

x=-1\pm i\sqrt{3}

Consider the following quadratic equation:

ax^2+bx+c=0

The solutions are found using the quadratic formula:

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

Let's find the roots of:

P\left(x\right)=x^2+x+5

We use:

• a=1
• b=1
• c=5

This gives:

x=\dfrac{-1\pm\sqrt{1^2-4\left(1\right)\left(5\right)}}{2\cdot1}

x=\dfrac{-1\pm\sqrt{-19}}{2}

x=\dfrac{-1}{2}\pm \dfrac{i\sqrt{19}}{2}

Consider the following quadratic equation:

ax^2+bx+c=0

The solutions are found using the quadratic formula:

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

Let's find the roots of:

P\left(x\right)=2x^2+5x+13

We use:

• a=2
• b=5
• c=13

This gives:

x=\dfrac{-5\pm\sqrt{5^2-4\left(2\right)\left(13\right)}}{2\cdot2}

x=\dfrac{-5\pm\sqrt{-79}}{4}

x=\dfrac{-5}{4} \pm \dfrac{i\sqrt{79}}{4}

Consider the following quadratic equation:

ax^2+bx+c=0

The solutions are found using the quadratic formula:

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

Let's find the roots of:

P\left(x\right)=x^2-3x+5

We use:

• a=1
• b=-3
• c=5

This gives:

x=\dfrac{-\left(-3\right)\pm\sqrt{\left(-3\right)^2-4\left(1\right)\left(5\right)}}{2\cdot1}

x=\dfrac{3\pm\sqrt{-11}}{2}

x=\dfrac{3\pm i \sqrt{11}}{2}

x=\dfrac{3}{2}\pm \dfrac{i\sqrt{11}}{2}

Consider the following quadratic equation:

ax^2+bx+c=0

The solutions are found using the quadratic formula:

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

Let's find the roots of:

P\left(x\right)=3x^2-2x+5

We use:

• a=3
• b=-2
• c=5

This gives:

x=\dfrac{-\left(-2\right)\pm\sqrt{\left(-2\right)^2-4\left(3\right)\left(5\right)}}{2\cdot3}

x=\dfrac{2\pm\sqrt{-56}}{6}

x=\dfrac{2}{6}\pm \dfrac{2i\sqrt{14}}{6}

x=\dfrac{1}{3}\pm \dfrac{i\sqrt{14}}{3}

Consider the following quadratic equation:

ax^2+bx+c=0

The solutions are found using the quadratic formula:

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

Let's find the roots of:

P\left(x\right)=4x^2-4x+16

We use:

• a=4
• b=-4
• c=16

This gives:

x=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4\left(4\right)\left(16\right)}}{2\cdot4}

x=\dfrac{4\pm\sqrt{-15\left(16\right)}}{8}

x=\dfrac{4\pm 4i\sqrt{15}}{8}

x=\dfrac{1}{2}\pm\dfrac{ i\sqrt{15}}{2}