Question

can someone help me with steps

0>x^2 +5x-2

Answer 1

0 > x^2 + 5x - 2
We can also rewrite this as
x^2 + 5x - 2 < 0
First thing to do is find the roots of the equation. Since the quadratic equation is not factorable in just one look, we use the quadratic formula:
x = ( -b +/- sqrt(b^2 - 4ac) ) / 2a
Therefore,
x = ( -5 +/- sqrt((-5)^2 - 4(1)(-2)) ) / 2(1)
x = ( -5 +/- sqrt(25 + 8) ) / 2
x = ( -5 +/- sqrt(33) ) / 2
In decimal,
x1 = 0.3729
x2 = -5.3729
Now to check whether the solution lies between these values of x or outside these boundaries, we get a value between 0.3729 and -5.3729 and check if it satisfies the equation. If it is, then the solution is the all values between x1 and x2, otherwise, it's outside.
For instance, we get x = 0 (zero is between the x1 and x2 values), thus
0 > x^2 + 5x - 2
0 > 0^2 + 5(0) - 2
0 > -2
Indeed zero is greater than -2. Therefore, the solution for x is -5.3729 < x < 0.3729

Hope this helps~ :)

Answer 2

You can also note that since the parabola opens upward,

x^2 + 5x - 2 < 0

will be the region between the roots. Don't be afraid to use what you know, even if it's geometric, not algebraic.