Question

Select the values that are solutions to the inequality x2 + 3x – 4 > 0.

Answer 1

Answer:

So any number in the following set is a solution:

$(-\infty,-4) \cup (1,\infty)$

given the inequality to solve was:

$x^2+3x-4>0$

Step-by-step explanation:

The left hand side is a quadratic while the right hand side is 0.

Since this is a quadratic>0, I'm going to factor the quadratic if possible and then solve that quadratic=0 for x.

That is I'm going to solve:

$x^2+3x-4=0$

Since a=1, I get to ask what multiplies to be c (-4) and add up to be b(3).

Those numbers are 4 and -1.

So the factored form for the equation is:

$(x+4)(x-1)=0$

Setting both factors equal to 0 since 0*anything=0:

x+4=0            and              x-1=0

 -4   -4                                 +1   +1

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x=-4              and                   x=1

Ok so if this wasn't a quadratic I would make a number line and choose numbers to plug into the quadratic to see which intervals would give me positive results.  I say positive due to the >0 part.

However since I know about the shapes of quadratics, I'm going to use that.

The quadratic function $f(x)=x^2+3x-4$ has x-intercepts (-4,0) and (1,0) and is open up.

I determine that it was opened up because the leading coefficient is 1 which is positive.

Now the left tail and right tail is what is above the x-axis so the solution set is:

$(-\infty,-4) \cup (1,\infty)$

Answer 2

Answer:

-6 and 5

Step-by-step explanation: