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Rom4ik [11]
3 years ago
15

When a quadratic equation has no linear term (variable raised to the first power), the solution becomes simpler. What are the tw

o possible solutions to 5x^2−15=0? Enter your two answers separated by a comma.
Mathematics
1 answer:
OLga [1]3 years ago
3 0

Answer: -\sqrt{3},\ +\sqrt{3}

Step-by-step explanation:

The given quadratic equation :_ 5x^2-15=0

To solve this , we first add 15 on the both the sides of the above equation, we get

5x^2-15+15=0+15\\\\\Rightarrow\ 5x^2=15

Now, we divide both sides by 5, we get

x^2=3

Taking square roots on both the sides , we get

x=\pm\sqrt{3}

Hence, the two possible solutions = -\sqrt{3},\ +\sqrt{3}

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Answer:

(3, 5)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtract Property of Equality

<u>Algebra I</u>

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Step-by-step explanation:

<u>Step 1: Define Systems</u>

y = 2x - 1

2x + 4y = 26

<u>Step 2: Solve for </u><em><u>x</u></em>

<em>Substitution</em>

  1. Substitute in <em>y</em>:                                                                                                2x + 4(2x - 1) = 26
  2. Distribute 4:                                                                                                    2x + 8x - 4 = 26
  3. Combine like terms:                                                                                       10x - 4 = 26
  4. [Addition Property of Equality] Add 4 on both sides:                                  10x = 30
  5. [Division Property of Equality] Divide 10 on both sides:                              x = 3

<u>Step 3: Solve for </u><em><u>y</u></em>

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  2. Substitute in <em>x</em>:                                                                                                y = 2(3) - 1
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