How do you use the quadratic formula to solve #-x^2+1=-6x^2-x#?

1 Answer
Mar 29, 2017

Answer:

Sub in coefficients from standard form into QF. Solve for #x#. In this case, there are no solutions.

Explanation:

So first off, we have to transform the equation to standard form.

#f(x)=-x^2+1=-6x^2-x#

To do this, we have to bring all the terms to one side and equate the equation to 0.

#f(x)=-x^2 + 6x^2+x+1=0#

Now we add like terms.

#f(x)=5x^2+x+1=0#

Once we have our equation in standard form, we use the quadratic formula: #x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}#.

We sub in the #a#, #b#, and #c# values. Not the variables - #x#.

#x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}#

#x=\frac{-(1)\pm\sqrt{(1)^2-4(5)(1)\ }}{2(5)}#

#x=\frac{-1\pm\sqrt{-19\ }}{10}#

I stopped here, because under the radical, we have a negative number. It's impossible to root a negative number - it becomes undefined.

The value under the radical indicates the number of zeros the equation has. A positive number greater than #0#, means there are two zeros. A value of #0# indicates there is only 1 zero. If there is a negative number under the radical, it means there are no zeros.

Therefore, the solution to this equation #f(x)=-x^2+1=-6x^2-x#, is that there are none. Without zeros, there are no solutions.

Hope this helps :)