F(x)=-x^2-3x+5-এর সমাধান করুন

$$F ( x ) = - x ^ { 2 } - 3 x + 5$$

Factor

$$-\left(x-\frac{-\sqrt{29}-3}{2}\right)\left(x-\frac{\sqrt{29}-3}{2}\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-x^{2}-3x+5=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 5}}{2\left(-1\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 5}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 5}}{2\left(-1\right)}$$

Multiply 4 times 5.

$$x=\frac{-\left(-3\right)±\sqrt{9+20}}{2\left(-1\right)}$$

Add 9 to 20.

$$x=\frac{-\left(-3\right)±\sqrt{29}}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{29}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±\sqrt{29}}{-2}$$

Now solve the equation x=\frac{3±\sqrt{29}}{-2} when ± is plus. Add 3 to \sqrt{29}.

$$x=\frac{\sqrt{29}+3}{-2}$$

Divide 3+\sqrt{29} by -2.

$$x=\frac{-\sqrt{29}-3}{2}$$

Now solve the equation x=\frac{3±\sqrt{29}}{-2} when ± is minus. Subtract \sqrt{29} from 3.

$$x=\frac{3-\sqrt{29}}{-2}$$

Divide 3-\sqrt{29} by -2.

$$x=\frac{\sqrt{29}-3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3-\sqrt{29}}{2} for x_{1} and \frac{-3+\sqrt{29}}{2} for x_{2}.

$$-x^{2}-3x+5=-\left(x-\frac{-\sqrt{29}-3}{2}\right)\left(x-\frac{\sqrt{29}-3}{2}\right)$$

Evaluate

$$5-3x-x^{2}$$