Hello, Having trouble with this one...
Solve the inequality:
(2x-3)(4x)+x^2=>(3x+4)^2
I did 2x-3*4x
9x^2-12x on the left side
on the right side:
9x^2+24x+16
I substracted everything on the right side over to the left and made it => zero
9x^2-12x-9x^2-24x-16=>0
-36x-16=>0
x=<4/-9
I'm not sure if this is right or if I made an error somewhere.
it is correct.
THANKS!
To solve the inequality (2x-3)(4x)+x^2 >= (3x+4)^2, you correctly expanded the expressions on both sides of the inequality sign. However, there seems to be an error when you subtracted everything on the right side over to the left side. Let's go through the correct steps to solve the inequality:
First, expand the expressions on both sides:
(2x-3)(4x) + x^2 >= (3x+4)^2
8x^2 - 12x + x^2 >= 9x^2 + 24x + 16
Combine like terms:
9x^2 - 12x >= 9x^2 + 24x + 16
Subtract 9x^2 from both sides:
-12x >= 24x + 16
Now we want to isolate the variable x on one side. To do that, let's move all the terms with x to the left side by subtracting 24x from both sides:
-12x - 24x >= 16
Combine like terms:
-36x >= 16
Next, we want to isolate x by dividing both sides of the inequality by -36. Since we divide by a negative number, we need to flip the direction of the inequality sign:
x <= 16 / -36
Finally, simplify the fraction:
x <= -4/9
So, the correct solution to the inequality is x <= -4/9.