Lesson: The Quadratic Formula

Algebra - Lesson Solving Quadratics with Quadratic Formula

This is the general form of a quadratic.

Remember, to solve this type of equation y must be equal to 0.

Multiple Choice

Identify the a, b and c values for this equation:

3x² - 8x + 15 = 0

• a = 3, b = 8, c = 0

• a = 3, b = -8, c = 15

• a = 3, b = 8, c = 15

Dropdown

Given the Quadratic Equation: 2x² + 5x + 3 = 0.

Identify the values to use in the Quadratic Formula.

a = ​

Select answer

^{b =} ​

Select answer

c = ​

Select answer

Dropdown

Given the Quadratic Equation: -16t² + 48t + 8 = 0.

Identify the values to use in the Quadratic Formula.

a = ​ ​ ​

Select answer

^{b =} ​ ​ ​

Select answer

c = ​ ​ ​

Select answer

Dropdown

Given the Quadratic Equation: -x² - 7x - 13 = 0.

Identify the values to use in the Quadratic Formula.

a = ​ ​

Select answer

^{b =} ​ ​

Select answer

c = ​ ​

Select answer

Multiple Choice

Which shows proper substitution into the Quadratic Formula to solve the equation:

$$x^2-7x+6=0$$

• $$x=\frac{-7\pm\sqrt[]{\left(-7\right)^2-4\left(1\right)\left(6\right)}}{2\left(1\right)}$$

• $$x=\frac{-\left(-7\right)\pm\sqrt[]{\left(-7\right)^2-4\left(1\right)\left(6\right)}}{2\left(1\right)}$$

• $$x=\frac{-\left(-7\right)\pm\sqrt[]{-7^2-4\left(1\right)\left(6\right)}}{2\left(1\right)}$$

• $$x=\frac{-7\pm\sqrt[]{-7^2-4\left(1\right)\left(6\right)}}{2\left(1\right)}$$

Multiple Choice

Use the Quadratic Formula to solve this equation. Simplify answers.

$$2x^2+x-15=0$$

• $$x=\frac{-\left(1\right)\pm\sqrt[]{\left(1\right)^2-4\left(2\right)\left(-15\right)}}{2\left(2\right)}$$ $$x=\frac{-1\pm\sqrt[]{121}}{4}$$

  $$x=\frac{-1\pm11}{4}$$

  $$x=\frac{-1-11}{4}$$

  or

  $$x=\frac{-1+11}{4}$$

  $$x=-\frac{12}{4}\ or\ x=\frac{10}{4}$$

  $$x=-3\ \ or\ \ x=2.5$$

• $$x=\frac{-\left(1\right)\pm\sqrt[]{\left(1\right)^2-4\left(2\right)\left(15\right)}}{2\left(2\right)}$$

  $$x=\frac{-1\pm\sqrt[]{-119}}{4}$$

  No real solution since the square root of -119 is NOT a real number.

• $$x=\frac{-\left(1\right)\pm\sqrt[]{\left(1\right)^2-4\left(2\right)\left(-15\right)}}{2\left(2\right)}$$ $$x=\frac{-1\pm\sqrt[]{121}}{4}$$

  $$x=\frac{-1\pm11}{4}$$

  $$x=\frac{-1-11}{4}$$

  or

  $$x=\frac{-1+11}{4}$$

  $$x=-\frac{12}{4}\ or\ x=\frac{10}{4}$$

  $$x=3\ \ or\ \ x=-2.5$$

• $$x=\frac{-\left(1\right)\pm\sqrt[]{\left(1\right)^2-4\left(2\right)\left(-15\right)}}{2\left(2\right)}$$ $$x=\frac{-1\pm\sqrt[]{121}}{4}$$

  $$x=\frac{-1\pm11}{4}$$

  $$x=\frac{-1-11}{4}$$

  or

  $$x=\frac{-1+11}{4}$$

  $$x=\frac{12}{4}\ or\ x=\frac{10}{4}$$

  $$x=3\ \ or\ \ x=2.5$$

Multiple Choice

Use the quadratic formula to find the solutions for

y = -x² - 5x + 12

• No Real Solution

Multiple Choice

What would the equation be the a, b coefficients, and the constant c?

*remember, it should = ZERO before you identify a, b & c!

$$3p^2-2p+19\ =\ 24$$

• a= 3, b= -2, c= 19

• a= 19, b= 3, c= -2

• a= 3, b= -2, c= -5

• a= 3, b= -2, c= 43

Multiple Choice

Which correctly shows the proper use of the quadratic formula for the equation? $$3p^2-2p-5$$

• $$\frac{-\left(3\right)\pm\sqrt[]{\left(-5\right)^2-4\left(3\right)\left(-2\right)}}{2\left(-2\right)}$$

• $$\frac{-\left(2\right)\pm\sqrt[]{\left(-2\right)^2-4\left(3\right)\left(-5\right)}}{2\left(3\right)}$$

• $$\frac{-\left(-2\right)\pm\sqrt[]{\left(3\right)^2-4\left(3\right)\left(-2\right)}}{2\left(-5\right)}$$

• $$\frac{-\left(-2\right)\pm\sqrt[]{\left(-2\right)^2-4\left(3\right)\left(-5\right)}}{2\left(3\right)}$$

Multiple Choice

What are the solutions to the equation? $$-8r^2-9r-1=-9$$

• $$-\frac{9\pm\sqrt[]{337}}{16}$$

• $$-\frac{5\pm2\ \sqrt[]{52}}{18}$$

• $$-\frac{5\pm\sqrt[]{176}}{18}$$

• $$-\frac{2\pm\sqrt[]{337}}{9}$$