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

Mar 29, 2017

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.

$f \left(x\right) = - {x}^{2} + 1 = - 6 {x}^{2} - x$

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

$f \left(x\right) = - {x}^{2} + 6 {x}^{2} + x + 1 = 0$

Now we add like terms.

$f \left(x\right) = 5 {x}^{2} + x + 1 = 0$

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

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

$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c \setminus}}{2 a}$

$x = \setminus \frac{- \left(1\right) \setminus \pm \setminus \sqrt{{\left(1\right)}^{2} - 4 \left(5\right) \left(1\right) \setminus}}{2 \left(5\right)}$

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

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 $f \left(x\right) = - {x}^{2} + 1 = - 6 {x}^{2} - x$, is that there are none. Without zeros, there are no solutions.

Hope this helps :)