How do you solve # 14-5x-x^2

Question

How do you solve # 14-5x-x^2<0#?

Eli Assouline · Accepted Answer

#x<-7 uu x>2#

Just to make things easier, we multiply both sides by #-1# and flip the inequality sign: #x^2+5x-14>0#. This way the coefficient before #x^2# is positive. (Note: this is not a necessary step.)

We factorize the expression: #x^2+5x-14=(x+7)(x-2)>0# to find the roots: #x=-7# or #2#.

These are the only two possible points where the expression can change signs. Thus, there are three regions that have the same sign:

Since there are no repeated roots, the signs of the region alternate. We substitute an arbitrary number in the second region to find the sign: #x=0#. Then #x^2+5x-14=0^2+5*0-14=-14<0#. Thus, the second region is negative.

Therefore, the first and third regions are positive (you can use substitution to be sure). Looking at the inequality, #x^2+5x-14>0#, we notice that we need to find regions where the value is positive.

The solution is the first and third regions: #x<-7 uu x>2#.

Abigail Allen · Answer

undefined To solve $ 14 - 5x - x^2 < 0 $:

1. Rearrange the equation to set it equal to zero: $ -x^2 - 5x + 14 < 0 $.
2. Factor the quadratic expression if possible.
3. Find the critical points by setting the expression equal to zero.
4. Use test points in each interval determined by the critical points to determine where the inequality holds true.
5. Write the solution set based on the intervals where the inequality is satisfied.

Alternatively, you can use the quadratic formula to find the roots of the quadratic equation and then analyze the sign of the expression within the intervals determined by the roots.

How do you solve # 14-5x-x^2&lt;0#?

How do you solve # 14-5x-x^2<0#?