How do you solve #-s^2+4s-6

Question

How do you solve #-s^2+4s-6<0#?

James Allen · Accepted Answer

Solution: # x in RR # or #(-oo,oo) #

# -s^2 +4s -6 < 0 a= -1 ,b = 4 , c= -6 #

Vertex #(x) = -b/2a= -4/-2= 2# .

Vertex #(y) = - 2^2+4*-2 -6= -4+8-6 =-2#

Vertex is # 2,-2 # . The parabola opens down wards since #a <0#

The range is #< -2 # i.e #<0# . The domain is any real value

i.e # x in RR # or #(-oo,oo)# . The graph also shows

#- s^2 +4s -6 # is # < -2 # i.e # < 0# .

Solution: # x in RR # or #(-oo,oo) #

graph{-x^2+4x-6 [-12.66, 12.65, -6.33, 6.33]}

Liam Brennan · Answer

undefined To solve the inequality $ -s^2 + 4s - 6 < 0 $, follow these steps:

1. Start by rearranging the inequality so that the quadratic expression is on one side and zero is on the other side:

$$ -s^2 + 4s - 6 < 0 $$
   $$ -s^2 + 4s - 6 = 0 $$

2. Next, solve the quadratic equation $ -s^2 + 4s - 6 = 0 $ to find the critical points. You can use the quadratic formula or factorization to solve for $ s $. The solutions are the points where the quadratic equation crosses the x-axis.

3. Once you have the critical points, which are the solutions to $ -s^2 + 4s - 6 = 0 $, plot them on a number line. These points divide the number line into intervals.

4. Test a value from each interval in the original inequality $ -s^2 + 4s - 6 < 0 $. Determine whether the inequality is true or false for each interval.

5. The solution to the inequality is the set of values of $ s $ that make the inequality true. This will be the interval(s) where the inequality is satisfied based on your testing in step 4.

By following these steps, you can find the solution to the inequality $ -s^2 + 4s - 6 < 0 $ and determine the range of values for $ s $ that satisfy the inequality.

How do you solve #-s^2+4s-6&lt;0#?

How do you solve #-s^2+4s-6<0#?