Solving Quadratic Equations with Substitution

The video tutorial explains how to solve a quadratic equation using substitution. By substituting a part of the equation with a variable 'u', the equation is simplified into a basic quadratic form. The tutorial then demonstrates factoring the equation and solving for the variable 'y'. Finally, it verifies the solutions by substituting them back into the original equation to ensure they are correct.

What type of equation is being solved in the video?

• Exponential equation

• Cubic equation

• Quadratic equation

• Linear equation

What substitution is made to simplify the equation?

• u = y^2 + 5

• u = 3y - 5

• u = 3y + 5

• u = y + 3

What is the equation after setting it to zero and simplifying?

• u^2 + 3u - 10 = 0

• u^2 + 3u + 10 = 0

• u^2 - 3u - 10 = 0

• u^2 - 3u + 10 = 0

What are the factors of the equation u^2 + 3u - 10 = 0?

• (u - 2)(u + 5)

• (u - 3)(u + 3)

• (u + 2)(u - 5)

• (u + 2)(u + 5)

What is the first step in rewriting the equation back in terms of y?

• Multiply both sides by 3

• Divide both sides by 5

• Substitute 3y + 5 for u

• Add 5 to both sides

What is the solution for y when 3y + 3 = 0?

• y = 0

• y = -1

• y = 3

• y = 1

What is the solution for y when 3y + 10 = 0?

• y = 3

• y = -3

• y = 10/3

• y = -10/3

Why is substitution used in solving the equation?

• To make the equation more complex

• To avoid squaring the binomial

• To eliminate the variable y

• To increase the degree of the equation

What is verified by substituting the solutions back into the original equation?

• The equation has no solutions

• The equation is linear

• The solutions satisfy the original equation

• The solutions are incorrect

What is the result when y = -1 is substituted back into the original equation?

• 7 = 7

• 4 = 4

• 6 = 6

• 5 = 5