Solve Using the Quadratic Formula 2-x^2-x=0

Solution:

Step 1:

Apply the quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Step 2:

Insert the coefficients $a=-1$, $b=-1$, and $c=2$ into the formula: $x=\frac{-(-1)\pm \sqrt{(-1{)}^2-4(-1)(2)}}{2(-1)}$.

Step 3:

Commence simplification of the expression.

Step 3.1:

Begin by simplifying the expression in the numerator.

Step 3.1.1:

Calculate the square of $-1$: $x=\frac{1\pm \sqrt{1+(-4)(-1)(2)}}{-2}$.

Step 3.1.2:

Perform the multiplication inside the square root.

Step 3.1.2.1:

Calculate $-4$ times $-1$: $x=\frac{1\pm \sqrt{1+4(2)}}{-2}$.

Step 3.1.2.2:

Multiply $4$ by $2$: $x=\frac{1\pm \sqrt{1+8}}{-2}$.

Step 3.1.3:

Add $1$ to $8$: $x=\frac{1\pm \sqrt{9}}{-2}$.

Step 3.1.4:

Express $9$ as a square of $3$: $x=\frac{1\pm 3}{-2}$.

Step 3.1.5:

Extract square root values, assuming they are positive real numbers: $x=\frac{1\pm 3}{-2}$.

Step 3.2:

Multiply the denominator $2$ by $-1$: $x=\frac{1\pm 3}{-2}$.

Step 3.3:

Place the negative sign in front of the fraction: $x=-\frac{1\pm 3}{2}$.

Step 4:

Combine both solutions to get the final answer: $x=-2,1$.

Knowledge Notes:

The quadratic formula is a fundamental tool for solving quadratic equations of the form $a{x}^2+bx+c=0$. The formula is given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the equation, and $\sqrt{b^2-4ac}$ is called the discriminant. The discriminant determines the nature of the roots: if it is positive, there are two real and distinct solutions; if it is zero, there is one real solution; and if it is negative, there are two complex solutions.

In the given problem, the equation is $2-x^2-x=0$. To match the standard form $a{x}^2+bx+c=0$, we rewrite the equation as $-x^2-x+2=0$. Here, $a=-1$, $b=-1$, and $c=2$. We substitute these values into the quadratic formula to find the solutions for $x$.

The process involves arithmetic operations such as squaring numbers, multiplying, adding, and simplifying square roots. When simplifying square roots of perfect squares, we obtain integer values. After simplifying the numerator and denominator, we obtain the solutions for $x$.

The final step is to present both possible solutions for $x$, which result from the "$\pm$" in the quadratic formula. These solutions are the roots of the original quadratic equation.