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Solve by completing the square.

-2x2+5x+1=0

Aug 31, 2017

#1
+2339
+2

I actually do not like the method of completing the square because I believe it to be quite inefficient, but I'll use it, if you insist.

Using the method of completing the square, I will solve for x in the equation $$-2x^2+5x+1=0$$.

$$-2x^2+5x+1=0$$ In a quadratic equation in the form, $$ax^2+bx+c=0$$ we must move c to the other side; in other words, subtract 1 on both sides of the equation.
$$-2x^2+5x=-1$$ Before we can proceed, we must make the coefficient of the quadratic term 1. At the moment, it is -2. Therefore, we must divide by -2 on both sides.
$$x^2-\frac{5}{2}x=\frac{1}{2}$$ Now, the next part is probably the hardest to understand. Our goal in the next step is make $$x^2-\frac{5}{2}x$$ a perfect-square trinomial by adding a value. In order to do that, we must add $$\left(\frac{b}{2}\right)^2$$ where b is the coefficient of the linear term.
$$x^2-\frac{5}{2}x+\left(\frac{b}{2}\right)^2=\frac{1}{2}+\left(\frac{b}{2}\right)^2$$ I added (b/2)^2 to both sides because whatever you do one side, you must do to the other in order to keep the equations balanced. Now, plug in the appropriate value of b, (-5/2) in this case.
$$x^2-\frac{5}{2}x+\left(\frac{\frac{-5}{2}}{2}\right)^2=\frac{1}{2}+\left(\frac{\frac{-5}{2}}{2}\right)^2$$ Now, I will simplify within the parentheses.
$$\frac{\left(\frac{-5}{2}\right)}{2}=\frac{-5}{2}\div\frac{2}{1}=\frac{-5}{2}*\frac{1}{2}=\frac{-5}{4}$$ Reinsert this.
$$x^2-\frac{5}{2}x+\left(\frac{-5}{4}\right)^2=\frac{1}{2}+\left(\frac{-5}{4}\right)^2$$ Distribute the exponent to both the numerator and denominator.
$$x^2-\frac{5}{2}x+\frac{(-5)^2}{4^2}=\frac{1}{2}+\frac{(-5)^2}{4^2}$$ Simplify both the numerator and the denominator of both fractions.
$$x^2-\frac{5}{2}x+\frac{25}{16}=\frac{1}{2}+\frac{25}{16}$$ Let's simplify the right hand side of the equation. In this case, this requires converting 1/2 into a fraction with the denominator of 16.
$$\frac{1}{2}+\frac{25}{16}=\frac{8}{16}+\frac{25}{16}=\frac{33}{16}$$
$$x^2-\frac{5}{2}x+\frac{25}{16}=\frac{33}{16}$$ By adding (b/2)^2 to both sides of the equation, we have created a perfect-square trinomial on the left-hand side. It may not be obvious because of the fractions, but it may help to know that the trinomial, when manipulated correctly, will always become in the form of $$\left(x+\frac{b}{2}\right)^2$$
$$\left(x-\frac{5}{4}\right)^2=\frac{33}{16}$$ Take the square root of both sides. Remember that the square root of both sides results in both the absolute value.
$$\left|x-\frac{5}{4}\right|=\sqrt{\frac{33}{16}}$$ Let's simplify the right hand side of the equation by distributing the square root to both the numerator and the denominator.
$$\left|x-\frac{5}{4}\right|={\frac{\sqrt{33}}{\sqrt{16}}}$$ The square root of 16 is 4, so let's simplify that.
$$\left|x-\frac{5}{4}\right|=\frac{\sqrt{33}}{4}$$ Now, solve for both positive and negative answer.
 $$x-\frac{5}{4}=\frac{\sqrt{33}}{4}$$ $$-\left(x-\frac{5}{4}\right)=\frac{\sqrt{33}}{4}$$

 $$x=\frac{\sqrt{33}}{4}+\frac{5}{4}$$ $$-x+\frac{5}{4}=\frac{\sqrt{33}}{4}$$

 $$x_1=\frac{\sqrt{33}+5}{4}$$ $$-x=\frac{\sqrt{33}}{4}-\frac{5}{4}$$

 $$x_1=\frac{\sqrt{33}+5}{4}$$ $$-x=\frac{\sqrt{33}-5}{4}$$

 $$x_1=\frac{\sqrt{33}+5}{4}$$ $$x_2=\frac{-\sqrt{33}+5}{4}$$

Aug 31, 2017

#1
+2339
+2

I actually do not like the method of completing the square because I believe it to be quite inefficient, but I'll use it, if you insist.

Using the method of completing the square, I will solve for x in the equation $$-2x^2+5x+1=0$$.

$$-2x^2+5x+1=0$$ In a quadratic equation in the form, $$ax^2+bx+c=0$$ we must move c to the other side; in other words, subtract 1 on both sides of the equation.
$$-2x^2+5x=-1$$ Before we can proceed, we must make the coefficient of the quadratic term 1. At the moment, it is -2. Therefore, we must divide by -2 on both sides.
$$x^2-\frac{5}{2}x=\frac{1}{2}$$ Now, the next part is probably the hardest to understand. Our goal in the next step is make $$x^2-\frac{5}{2}x$$ a perfect-square trinomial by adding a value. In order to do that, we must add $$\left(\frac{b}{2}\right)^2$$ where b is the coefficient of the linear term.
$$x^2-\frac{5}{2}x+\left(\frac{b}{2}\right)^2=\frac{1}{2}+\left(\frac{b}{2}\right)^2$$ I added (b/2)^2 to both sides because whatever you do one side, you must do to the other in order to keep the equations balanced. Now, plug in the appropriate value of b, (-5/2) in this case.
$$x^2-\frac{5}{2}x+\left(\frac{\frac{-5}{2}}{2}\right)^2=\frac{1}{2}+\left(\frac{\frac{-5}{2}}{2}\right)^2$$ Now, I will simplify within the parentheses.
$$\frac{\left(\frac{-5}{2}\right)}{2}=\frac{-5}{2}\div\frac{2}{1}=\frac{-5}{2}*\frac{1}{2}=\frac{-5}{4}$$ Reinsert this.
$$x^2-\frac{5}{2}x+\left(\frac{-5}{4}\right)^2=\frac{1}{2}+\left(\frac{-5}{4}\right)^2$$ Distribute the exponent to both the numerator and denominator.
$$x^2-\frac{5}{2}x+\frac{(-5)^2}{4^2}=\frac{1}{2}+\frac{(-5)^2}{4^2}$$ Simplify both the numerator and the denominator of both fractions.
$$x^2-\frac{5}{2}x+\frac{25}{16}=\frac{1}{2}+\frac{25}{16}$$ Let's simplify the right hand side of the equation. In this case, this requires converting 1/2 into a fraction with the denominator of 16.
$$\frac{1}{2}+\frac{25}{16}=\frac{8}{16}+\frac{25}{16}=\frac{33}{16}$$
$$x^2-\frac{5}{2}x+\frac{25}{16}=\frac{33}{16}$$ By adding (b/2)^2 to both sides of the equation, we have created a perfect-square trinomial on the left-hand side. It may not be obvious because of the fractions, but it may help to know that the trinomial, when manipulated correctly, will always become in the form of $$\left(x+\frac{b}{2}\right)^2$$
$$\left(x-\frac{5}{4}\right)^2=\frac{33}{16}$$ Take the square root of both sides. Remember that the square root of both sides results in both the absolute value.
$$\left|x-\frac{5}{4}\right|=\sqrt{\frac{33}{16}}$$ Let's simplify the right hand side of the equation by distributing the square root to both the numerator and the denominator.
$$\left|x-\frac{5}{4}\right|={\frac{\sqrt{33}}{\sqrt{16}}}$$ The square root of 16 is 4, so let's simplify that.
$$\left|x-\frac{5}{4}\right|=\frac{\sqrt{33}}{4}$$ Now, solve for both positive and negative answer.
 $$x-\frac{5}{4}=\frac{\sqrt{33}}{4}$$ $$-\left(x-\frac{5}{4}\right)=\frac{\sqrt{33}}{4}$$

 $$x=\frac{\sqrt{33}}{4}+\frac{5}{4}$$ $$-x+\frac{5}{4}=\frac{\sqrt{33}}{4}$$

 $$x_1=\frac{\sqrt{33}+5}{4}$$ $$-x=\frac{\sqrt{33}}{4}-\frac{5}{4}$$

 $$x_1=\frac{\sqrt{33}+5}{4}$$ $$-x=\frac{\sqrt{33}-5}{4}$$

 $$x_1=\frac{\sqrt{33}+5}{4}$$ $$x_2=\frac{-\sqrt{33}+5}{4}$$

TheXSquaredFactor Aug 31, 2017