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ACT Math Intermediate Algebra Drill 2, Problem 2. How else can the expression be written?
ACT Math: Intermediate Algebra Drill 2, Problem 3. Which of the following is equal to the expression shown?
ACT Math: Intermediate Algebra Drill 4, Problem 2. What is the root of this equation?
ACT Math 4.2 Intermediate Algebra 433 Views
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ACT Math: Intermediate Algebra Drill 4, Problem 2. What is the root of this equation?
- Intermediate Algebra / Roots of polynomials
- Intermediate Algebra / Quadratic equations (Quadratic Formula)
- Product Type / ACT Math
- Foreign Language / Arabic Subtitled
- Foreign Language / Korean Subtitled
- Foreign Language / Chinese Subtitled
- Foreign Language / Spanish Subtitled
- Intermediate Algebra / Quadratic formula
Transcript
- 00:02
And here is your shmoop de jour...
- 00:06
What is a root of this equation? y2 - y = 4
- 00:10
And here are the potential answers...
- 00:16
OK so what is this question asking?
- 00:18
Sneaky quadratic equation question. We need to pull the root numbers and plug them in
Full Transcript
- 00:25
and see what we get. Remember the quadratic equation? Our old buddy,
- 00:29
old pal? When an equation is in the form ax squared
- 00:33
plus bx plus c equals 0... ...the solution can be negative b plus or
- 00:39
minus radical, b-squared minus 4ac all over 2a.
- 00:45
That's the quadratic equation formula.
- 00:47
That's a mouthful. Let's change the equation to y squared minus y minus
- 00:57
4 equals 0... so it looks like ax squared plus bx plus c equals 0.
- 01:06
We can see that a is 1, b is negative 1, and c is negative 4.
- 01:11
When we start pluggin', we get that 1 plus or minus the square root of 1 minus 4 times
- 01:22
1 times negative 4 all over 2... ...gives us 1 plus or minus the square root
- 01:29
of 17 over 2. So now we have one solution, or root, of the equation.
- 01:34
Looks like D.
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