Mathematics, 13.12.2020 19:20 potternatalie90

Use the quadratic formula to solve for x. 3x2 = 8x-1 Round your answer to the nearest hundredth. If there is more than one solution, separate them with commas. = 0 X $ ?

Answers: 3

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Answer from: ultimatesaiyan

$x \approx 0.13, 2.54$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Algebra I

Standard Form: ax² + bx + c = 0Quadratic Formula: $x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Step-by-step explanation:

Step 1: Define

3x² = 8x - 1

Step 2: Rewrite

Put equation in Standard Form

Subtract 8x on both sides: 3x² - 8x = -1Add 1 to both sides: 3x² - 8x + 1 = 0

Step 3: Identify Variables

a = 3

b = -8

c = 1

Step 4: Solve for x

Substitute [Quad Formula]: $x=\frac{8\pm\sqrt{(-8)^2-4(3)(1)} }{2(3)}$ Evaluate Exponents: $x=\frac{8\pm\sqrt{64-4(3)(1)} }{2(3)}$ Multiplication: $x=\frac{8\pm\sqrt{64-12} }{6}$ Subtraction: $x=\frac{8\pm\sqrt{52} }{6}$ Simplify Radical: $x=\frac{8\pm2\sqrt{13} }{6}$ Factor GCF: $x=\frac{2(4\pm\sqrt{13}) }{6}$ Simplify Fraction: $x=\frac{4\pm\sqrt{13}}{3}$ Evaluate: x=0.131483, 2.53518

Round: $x \approx 0.13, 2.54$

Answer from: Quest

7.25

step-by-step explanation:

we know that 7 equals to 7.0. we also know that 1/4=0.25. so we can add the two together and we get 7.25

Answer from: Quest

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly, giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.

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Use the quadratic formula to solve for x. 3x2 = 8x-1 Round your answer to the nearest hundredth. If...