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2 years ago

Solving these quadratics for x using complete the square 3x^2-4x-1=0. Show your work.

1 answer:

6 0

Answer: x = (sqrt(7) + 2)/3 and
x = ( – sqrt(7) + 2)/3

Explanation:

3x^2 - 4x - 1 = 0

Divide both sides by 3:

3x^2/3 - 4/3x - 1/3 = 0/3
x^2 - 4/3x - 1/3 = 0
x^2 - 4/3x = 1/3
x^2 - 4/3x + (2/3)^2 = 1/3 + (2/3)^2
(x - 2/3)^2 = 1/3 + 4/9
(x - 2/3)^2 = 7/9

Sqrt both sides:

x - 2/3 = sqrt (7/9)
x - 2/3 = |sqrt(7)/3|

Set x -2/3 = sqrt(7)/3

=> x = (sqrt(7) + 2)/3

Set x - 2/3 = - sqrt(7)/3

=> x = ( - sqrt(7) + 2)/3

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Consider a sequence given by the formula f(n)=3n-1 starting with n=1. Generate the first 5 terms of the sequence

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The first 5 terms are 2, 5, 8, 11, 14.

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The answer to the given question is as below:-

Polynomials:-

x³-7x²+9x-5x⁴-20

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Non-polynomial:-

x⁻5-5x⁻⁴+4x⁻³+2x⁻1-1

$\dfrac{4}{x^4}+ \dfrac{3}{x^3}-\dfrac{2}{x^4}-1$

$\sqrt[4]{x} -\sqrt[3]{x} 4+\sqrt{x}-8x+16$

<h3>What are polynomials?</h3>

A polynomial is a mathematical equation made up of indeterminates and coefficients and involves only addition, subtraction, multiplication, and non-negative integer exponentiation of variables. The exponents of the polynomials can not be irrational.

So the table will be formed as:-

Polynomials:- Non-polynomial:-

x³-7x²+9x-5x⁴-20 x⁻5-5x⁻⁴+4x⁻³+2x⁻1-1

x⁵-5x⁴+4x³+2x-1 $\dfrac{4}{x^4}+ \dfrac{3}{x^3}-\dfrac{2}{x^4}-1$

3x²-5x⁴+2x-12 $\sqrt[4]{x} -\sqrt[3]{x} 4+\sqrt{x}-8x+16$

Therefore the polynomials and non-polynomials are shown in the table above.

To know more about polynomials follow

brainly.com/question/2833285

#SPJ1

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