Let's solve the equation step by step:
Given equation:
[ 299x^2 + 100x = 500 - 101x^2 ]
First, let's move all terms to one side of the equation:
[ 299x^2 + 100x + 101x^2 - 500 = 0 ]
Combine like terms:
[ (299x^2 + 101x^2) + 100x - 500 = 0 ]
[ 400x^2 + 100x - 500 = 0 ]
Now, we have a standard quadratic equation of the form:
[ ax^2 + bx + c = 0 ]
where ( a = 400 ), ( b = 100 ), and ( c = -500 ).
We can simplify this equation by dividing all terms by 100:
[ 4x^2 + x - 5 = 0 ]
Now, we can solve for ( x ) using the quadratic formula:
[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
Substituting in the values of ( a ), ( b ), and ( c ):
[ x = \frac{-1 \pm \sqrt{(1)^2 - 4 \cdot 4 \cdot (-5)}}{2 \cdot 4} ]
[ x = \frac{-1 \pm \sqrt{1 + 80}}{8} ]
[ x = \frac{-1 \pm \sqrt{81}}{8} ]
[ x = \frac{-1 \pm 9}{8} ]
Now, we calculate the two possible values for ( x ):
- ( x = \frac{-1 + 9}{8} = \frac{8}{8} = 1 )
- ( x = \frac{-1 - 9}{8} = \frac{-10}{8} = -\frac{5}{4} )
Thus, the solutions to the equation ( 299x^2 + 100x = 500 - 101x^2 ) are:
[ x = 1 \quad \text{and} \quad x = -\frac{5}{4} ]
