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How do you factor and solve #2x^2 + x - 1=0#?

Answer 1

Abigail Allen

#x=1/2# or #x=-1#

#2x^2+x-1=0#

Factorise.

#2x^2+2x-x-1=0#

#2x(x+1)-1(x+1)=0#

#(2x-1)(x+1)=0#

#2x-1=0# or #x+1=0#

#x=1/2# or #x=-1#

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Answer 2

Avery Alexander

To factor and solve the quadratic equation (2x^2 + x - 1 = 0), you can use the quadratic formula: (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = 2), (b = 1), and (c = -1). Substituting these values into the formula, you get: (x = \frac{{-1 \pm \sqrt{{1^2 - 4 \cdot 2 \cdot (-1)}}}}{{2 \cdot 2}}). Simplify this expression and you get (x = \frac{{-1 \pm \sqrt{{9}}}}{{4}}). This simplifies to (x = \frac{{-1 \pm 3}}{{4}}), which gives two possible solutions: (x = \frac{{-1 + 3}}{{4}} = \frac{1}{2}) and (x = \frac{{-1 - 3}}{{4}} = -1). Therefore, the solutions to the equation (2x^2 + x - 1 = 0) are (x = \frac{1}{2}) and (x = -1).

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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.