solve the inequallity x^2+2x<3?

Question

please show ur working. thanks alot!!

Answer 1

x^2+2x-3 <0
=> (x+3)(x-1) < 0

This condition is met when only one of the factors is negative and the other is positive.

=> x < 1 and x > -3 (more precisely -3 < x <1)
OR
x > 1 and x < -3 (which is absurd)

So the solution is -3 < x < 1

Answer 2

Solve:
x^2 +2x - 3 < 0

Given that:
3*(-1) = -3 and 3-1 = 2

then , x^2 +2x - 3 = (x-1)(x+3) < 0

(x-1)(x+3) < 0
There are two options
that
x-1<0 OR x+3>0
AND
x-1>0 OR x+3<0

So
x>-3 OR x<1
It is the set: x E (-3; 1)

x<-3 OR x>1
It is the set: x E 0 (empty set)

Then,
x E (-3; 1)
AND
x E 0
is
x E ( -3; 1)

The solution is the set:
x E ( -3; 1)

IT IS WORTH NOTING THAT x = -3 and
x = 1 ARE NOT SOLUTIONS AT ALL.

Let's test it:
Out of set by the left side:
if x = -4, then (-4-1)*(-4+3) = (-5)(-1) = 5 > 0
It does not sastify the inequation.

Out of set by the right side:
if x = 2, then (2-1)*(2+3) = (1)(5) = 5 > 0
It does not sastify the inequation.

Inside of the set:
if x = 0, then (0-1)*(0+3) = (-1)(3) = -3 < 0
Yes, It does sastify the inequation.

if x = 0.5, then (0.5-1)*(0.5+3) = (-0.5)(3.5) = -1.75 < 0
Yes, It does sastify the inequation.

On the edge of set [left]:
if x = -3, then (-3-1)*(-3+3) = (-4)(0) = 0
It does not sastify the inequation.

On the edge of set [right]:
if x = 1, then (1-1)*(1+3) = (0)(4) = 0
It does not sastify the inequation.

You're Welcome!
:)

Answer 3

x^2+2x<3 (start by writing down problem). Subtract 3 from both sides. x^2+2x-3<0. Factor left hand side, and solve for x. (x+3)(x-1)<0. x=-3 and x=1. These are your test points. Test for x=-4,x=-2, and x=2. Write down the range, which if you do it right will give you range of -3>x>1. I leave the work of testing the ranges to you.

Answer 4

add -3 on both sides
x^2+2x -3 < 0
or (x-1)(x+3) < 0

one is positive and one negative
1st -ve and second positive
x-1 < 0 and x+ 3 > 0 or x< 1 and x > -3 so < -3 < x < 1

if 1st positive and second -ve
x-1 > 0 and x + 3 <0 so x > 1 and x < -3 which is not a solution as contradiction

so -3 < x <1 is the solution;'

Answer 5

x^2 + 2X <3
X^2 +2X-3<0
(X+3)(x-1)<0
draw number line...if you sub values between -3 and 1 in the equation you'll find that its -ve while the values smaller than -3 and larger than 1 is +ve.
Hence
-3

Answer 6

Subtract 3 from both sides:
x^2+2x-3<0
(x+3)(x-1)<0
x+3<0
x-1<0
x>-3
x<1

The answer is -3

Answer 7

x^2 + 2x <3

assume, x^2 + 2x =3
or,
x^2 + 2x -3=0
or,
(x+ 3) (x -3)=0
x= - 3 , x=1
so, it becomes a parabola where f(x)=0 at these values

so, ans: (-3,1)

Answer 8

X^2 + 2x < 3

X^2 + 2x - 3 < 3

(x-1)(x+3) < 0

<---false----~3------------true-------------1---------false---------->

sorry number line didnt come out to well..


solution: (-3, 1)

.. my apologies if it is incorrect I only did it because I was bored and maybe you should do your own homework..