How do you factor and solve #x^2+4x-3=0#?

Answer 1

It doesn't factorise but it can still be solved, for example : By completing the square.

So let's complete the square.

#therefore# #x^2+4x-3 = (x+2)^2-4-3=0# #therefore (x+2)^2-7=0#

Now lets solve:

#(x+2)^2-7=0# #(x+2)^2=7# #x+2=+-sqrt(7)# #x=-2+-sqrt(7)#

Therefore X has been solved

#x=-2+sqrt(7)# or #x=-2-sqrt(7)#
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Answer 2

To factor and solve the quadratic equation (x^2 + 4x - 3 = 0), you can use the quadratic formula or factorization method.

Using the quadratic formula: [x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}] Where (a = 1), (b = 4), and (c = -3).

[x = \frac{{-4 \pm \sqrt{{4^2 - 4(1)(-3)}}}}{{2(1)}}] [x = \frac{{-4 \pm \sqrt{{16 + 12}}}}{{2}}] [x = \frac{{-4 \pm \sqrt{{28}}}}{{2}}] [x = \frac{{-4 \pm \sqrt{{4 \cdot 7}}}}{{2}}] [x = \frac{{-4 \pm 2\sqrt{7}}}{{2}}]

[x = -2 \pm \sqrt{7}]

So, the solutions are (x = -2 + \sqrt{7}) and (x = -2 - \sqrt{7}).

Alternatively, you can factor the quadratic expression: [x^2 + 4x - 3 = 0] [(x + 3)(x - 1) = 0]

Setting each factor equal to zero: (x + 3 = 0 \Rightarrow x = -3) (x - 1 = 0 \Rightarrow x = 1)

Therefore, the solutions are (x = -3) 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.

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.

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.

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.

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