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How do you solve #x^2 + 5x – 6 = 0 #?

Answer 1

Hazel Anglin

#color(blue)(x=1,x=-6# are the solutions for the expression.

#x^2+5x–6#

We can first factorise the above expression and then proceed towards finding the solution.

Factorising by splitting the middle term

#x^2+color(blue)(5x)–6 = x^2+color(blue)(6x - 1x)–6#

#=x(x+6) - 1(x +6)#

#color(blue)((x-1)(x+6) # is the factorised form for the expression. Now we can equate each factor with zero and obtain the solutions.

#x-1=0, color(blue)(x=1# #x+6=0, color(blue)(x=-6#

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Answer 2

Jameson Allen

To solve the equation (x^2 + 5x - 6 = 0), you can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where (a), (b), and (c) are the coefficients of the equation (ax^2 + bx + c = 0). In this case, (a = 1), (b = 5), and (c = -6). Plugging these values into the formula, we get:

(x = \frac{-5 \pm \sqrt{5^2 - 4 \times 1 \times -6}}{2 \times 1})

(x = \frac{-5 \pm \sqrt{25 + 24}}{2})

(x = \frac{-5 \pm \sqrt{49}}{2})

(x = \frac{-5 \pm 7}{2})

So, the solutions are (x = \frac{-5 + 7}{2} = 1) and (x = \frac{-5 - 7}{2} = -6).

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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.