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In terms of b, what is the solution to the quadratic equation x^2+by-4=0?

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In terms of b, what is the solution to the quadratic equation x^2+by-4=0?

Aug 12, 2017

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$$x^2+by-4=0\\ x=\dfrac{-b\pm\sqrt{b^2+16}}{2}$$

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Aug 12, 2017
#2
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I feel as if this was a careless oversight by you, but I think your question is what value for $$b$$makes $$x^2+bx-4=0$$. If this is not your question, then I apologize.

In a quadratic equation in the current form, $$ax^2+bx+c=0$$, the quadratic formula says that $$x = {-b \pm \sqrt{b^2-4ac} \over2a}$$. Now, let's solve for b:

 $$x = {-b \pm \sqrt{b^2-4ac} \over2a}$$ First, I will plug in the known values for a and c. $$x=\frac{-b\pm\sqrt{b^2-4(1)(-4)}}{2(1)}$$ Simplify inside the radical and the denominator $$x=\frac{-b\pm\sqrt{b^2+16}}{2}$$ Multiply both sides of the equation by 2. $$2x=-b\pm\sqrt{b^2+16}$$

At this point, there is a plus-minus symbol in the equation, so we will have to solve each equation separately:

 $$2x=-b+\sqrt{b^2+16}$$ Add b to both sides of the equation. $$2x+b=\sqrt{b^2+16}$$ Square both sides of the equation. Let's expand the left-hand side of the equation first. $$(2x+b)^2=(2x)^2+2(2x)(b)+b^2$$ Simplify this. $$4x^2+4xb+b^2$$ Now, let's expand the right side. $$(\sqrt{b^2+16})^2=b^2+16$$ Great, now that both sides are expanded. Let's reinsert both into their respective sides of the equation. $$4x^2+4xb+b^2=b^2+16$$ Subtract b2 from both sides. $$4x^2+4xb=16$$ Divide all sides of the equation by its GCF, 4. Doing this often ensures that your equation is in simplest form and doesn't have a common factor at the end. $$x^2+xb=4$$ Subtract x^2 from both sides. $$xb=4-x^2$$ Divide by x on both sides of the equation. $$b=\frac{4-x^2}{x}$$

Now that we have solved b for $$2x=-b+\sqrt{b^2+16}$$. Let's solve it when $$2x=-b-\sqrt{b^2+16}$$. It is simpler than it may seem, actually. You'll see why:

 $$2x=-b-\sqrt{b^2+16}$$ We are simply repeating the process from above. Add b to both sides. $$2x+b=-\sqrt{b^2+16}$$ Let's square both sides of the equation. $$(2x+b)^2=(-\sqrt{b^2+16})^2$$ When you square a number or expression, the negative and postive number are the same. In general, $$(-a)^2=a^2$$. $$(2x+b)^2=(\sqrt{b^2+16})^2$$ Does this equation seem familiar? It should! We just solved for it above. The solution for both are the same! This is because, when you solve, they become the same equation. We already know that $$b=\frac{4-x^2}{x}$$. There is no reason to go on.

This is the tediouis part, though. Because a square root is involved in this equation, we have to verify that  $$b=\frac{4-x^2}{x}$$ is actually a valid solution. Let's do that:

 $$x=\frac{-\left(\frac{4-x^2}{x}\right)+\sqrt{\left(\frac{4-x^2}{x}\right)^2+16}}{2}$$ First, let's get rid of the pesky fraction. Multiply both sides of the equation by 2 $$2x=-\left(\frac{4-x^2}{x}\right)+\sqrt{\left(\frac{4-x^2}{x}\right)^2+16}$$ Let's deal with the square root first. Let's do the menacing $$\left(\frac{4-x^2}{x}\right)^2$$. "Distribute" the exponent to both the numerator and denominator. $$\left(\frac{4-x^2}{x}\right)^2=\frac{(4-x^2)^2}{x^2}$$ Expand the numerator with the rule that $$(a-b^2=a^2-2ab-b^2)$$. $$(4-x^2)^2=4^2-2(4)(x^2)+(x^2)^2$$ Simplify this. $$16-8x^2+(x^2)^2$$ Knowing that$$(a^b)^c=a^{b*c}$$, simplify this. $$16-8x^2+x^4$$ Reinsert this back into the original equation. $$2x=-\left(\frac{4-x^2}{x}\right)+\sqrt{\frac{x^4-8x^2+16}{x^2}+16}$$ Add $$\frac{4-x^2}{x}$$ to both sides of the equation. $$2x+\frac{4-x^2}{x}=\sqrt{\frac{x^4-8x^2+16}{x^2}+16}$$ Now, we will have to square again! Let's do the left side first, since it will be difficult to do in just one step. $$\left(2x+\frac{4-x^2}{x}\right)^2=(2x)^2+2(2x)\left(\frac{4-x^2}{x}\right)+\left(\frac{4-x^2}{x}\right)^2$$ Let's simplify the easy bits first. $$(2x)^2=2x*2x=4x^2$$ Now, the harder one to simplify: 2*2x*((4-x^2)/(x^2)). First, distribute the 2 first. $$2(2x)\left(\frac{4-x^2}{x}\right)=\frac{4x}{1}*\frac{4-x^2}{x}$$ Before we do multiplication with the fractions, note that an x from the numerator of and the denominator of another can be canceled out, which simplifies the process significantly. $$\frac{4}{1}*\frac{{4-x^2}}{1}=16-4x^2$$ Now, normally simplifying$$\left(\frac{4-x^2}{x}\right)^2$$ would be difficult, but we already know what it equals since we already did it. $$\left(\frac{4-x^2}{x}\right)^2=\frac{x^4-8x^2+16}{x^2}$$ Reinsert all of these pieces back into the equation. $$\left(2x+\frac{4-x^2}{x}\right)^2=16+\frac{x^4-8x^2+16}{x^2}$$ In this case, the $$4x^2$$ and $$-4x^2$$ cancel out. Now, let's compare that with what is on the right-hand side of the equation. $$16+\frac{x^4-8x^2+16}{x^2}=\frac{x^4-8x^2+16}{x^2}+16$$ If you examine both sides, you will notice that both are the same! All this algebra demonstrates that $$b=\frac{4-x^2}{x}$$ and that it is not an extraneous solution.

Ok, after all of this algebra, I know that some of it will be confusing; it confuses me! If you have any questions, I suggest you comment with a follow-up question. Please be specific about where your question is because it may be difficult to help if your question is too vague, and I cannot understand which part is confusing to you. I tried to break this down as best as I could, but it may be hard to digest all this sometimes. And that's fine. Just ask!

Therefore, when using the quadratic equation and then solving for b, $$b=\frac{4-x^2}{x}$$. Ok, I am done now. As I aforementioned, if you have any questions, just ask!

Aug 12, 2017