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How do you solve #6x²=1-x#?

Answer 1

Genesis Allen

#" at "y=0 :" "x=-1/2" and "x=1/3#

Write as #" "y=6x^2+x-1#

Compare to the form #" "y=ax^2+bx+c#

#color(blue)("Using the formula: "x=(-b+-sqrt(b^2-4ac))/(2a))#

#=>x=(-1+-sqrt((-1)^2-4(6)(-1)))/(2(6))#

#x=(-1+-sqrt(25))/12" "->" "x=-1/12+-5/12#

#=>" at "y=0 :" "x=-1/2" and "x=1/3#

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Answer 2

Hazel Anglin

To solve the equation 6x² = 1 - x, follow these steps:

Move all terms to one side to set the equation to zero: 6x² + x - 1 = 0.
Use the quadratic formula: x = [-b ± √(b² - 4ac)] / (2a), where a = 6, b = 1, and c = -1.
Plug in the values: x = [-(1) ± √((1)² - 4(6)(-1))] / (2(6)).
Simplify the expression under the square root: √(1 + 24) = √25 = 5.
Substitute back into the formula: x = [-(1) ± 5] / 12.
Solve for both possible values of x: x₁ = (-1 + 5) / 12 = 4 / 12 = 1/3, and x₂ = (-1 - 5) / 12 = -6 / 12 = -1/2.

Therefore, the solutions to the equation are x = 1/3 and x = -1/2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.