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lbvjy [14]
2 years ago
14

Write the quadratic equation whose roots are 1 and -5, and whose leading coefficient is 3.

Mathematics
1 answer:
Anna35 [415]2 years ago
7 0

Step-by-step explanation:

they're is a neat little trick.

but first of all, "roots" means the 0 solutions of the equation. like

ax² + bx + c = 0

there are 2 solutions for x to make the whole expression equal 0.

these are the "roots".

now to the little trick :

when is a multiplication resulting in 0 ?

when at least one of the factors is 0.

and any quadratic expression can be written as multiplication of 2 factors. like

c(x - a)(x - b) = cx² - cax - cbx + cab

what are the "roots" or 0 points ?

either

c = 0

x - a = 0 | x = a

x - b = 0 | x = b

the leading coefficient = 3.

that means nothing else than c = 3.

root = 1 means (x - 1) is one factor.

root = -5 means (x + 5) is the other factor.

so, we have

3(x - 1)(x + 5) = 3x² + 12x - 15

and the equation is

3x² + 12x - 15 = 0

or

3x² + 12x = 15

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General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
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  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

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  • Terms/Coefficients
  • Exponential Rule [Rewrite]:                                                                              \displaystyle b^{-m} = \frac{1}{b^m}

<u>Calculus</u>

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Derivative Notation

Basic Power Rule:

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Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle y = \frac{3}{2x^2}<u />

\displaystyle \text{Point} \ (1, \frac{3}{2})

<u>Step 2: Differentiate</u>

  1. [Function] Rewrite [Exponential Rule - Rewrite]:                                            \displaystyle y = \frac{3}{2}x^{-2}
  2. Basic Power Rule:                                                                                             \displaystyle y' = -2 \cdot \frac{3}{2}x^{-2 - 1}
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  4. Rewrite [Exponential Rule - Rewrite]:                                                              \displaystyle y' = \frac{-3}{x^3}

<u>Step 3: Solve</u>

  1. Substitute in coordinate [Derivative]:                                                              \displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1^3}
  2. Evaluate exponents:                                                                                         \displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1}
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Unit: Derivatives

Book: College Calculus 10e

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