Why do I have to use a quadratic equation?

Question

I don't understand why this doesn't work:
Problem: $$x^{2}+x-2=0$$ Solution: $$x^{2}+x=2$$ $$x+x=\sqrt{2}$$ $$2x≈1.4$$ $$x≈1.4/2$$ $$x≈0.7$$ Checking: (Note: I made the calculation with more decimals than the amount I wrote here; that explain $0.7^{2}=0.5$) $$0.7^{2}+0.7-2≈0.5+0.7-2≈-0.79$$ This method doesn't work...it gives $-0.79$ and it must give 0.

Instead I must do:

Good Solution: $$\frac{-b±\sqrt{b^{2}-4ac}}{2a}=0$$ $$\frac{-1±\sqrt{1^{2}-4\times1\times(-2)}}{2\times 1}=0$$ $$\frac{-1±\sqrt{1+8}}{2}=0$$ $$\frac{-1±\sqrt{9}}{2}=0$$ $$\frac{-1±3}{2}=0$$ $$x_{1}=\frac{-1+3}{2}=\frac{2}{2}=1$$ $$x_{2}=\frac{-1-3}{2}=\frac{-4}{2}=-2$$ Checking: $$1^{2}+1-2=1+1-2=0$$ $$(-2)^{2}+(-2)-2=4-2-2=0$$ And that gives a good result.

1. I don't understand why my first attempt doesn't work.
2. Also, I don't understand why it has 2 possible results, $1$ and $-2$.

Answer 1

This is wrong :

$$x^2 + x = 2 \Rightarrow x + x = \sqrt{2}$$

The square root applies to the whole term :

$$x^2+x=2 \Rightarrow \sqrt{x^2+x} = \sqrt2$$

if $x^2 + x \geq 0$.

This is why your attempt is mistaken.

The equation $x^2 + x - 2=0$ has two roots (not two possible results, this is mistaken as a phrase), because of the square power. Explaining why this form means $2$ solutions, can be done in a lot of ways.

Graphically and geometrically, an equation of the form : $ax^2 + bx + c = 0$ describes a parabola, which due to its geometric form, if it has one solution, it needs to have another one, thus two roots, either on the real plane, or in the complex plane, based on the sign of its discriminant.

Alternatively, you can observe two numbers that satisfy the equation, by factorization : $(x-1)(x+2)=0$, which means that either for $x=1$ or $x=-2$ the equation/expression is satisfied.

Equations of the form $ax^2 + bx + c = 0$, which is the ones your post is interested in, always have a straight-forward solution and are of the most common in entry mathematics.

You can read up on some more stuff for the quadratic equation here.

Answer 2

This doesn't work because the step $$x^2+x=2 \implies x+x=\sqrt 2$$ is wrong.

You cannot take root only on one term in LHS, you will have to take square root of complete LHS. That is $\sqrt{x^2+x}$.

I think an example will make it easy for you to understand.

What you did : $$16+9=25 \implies \sqrt{16}+9=\sqrt{25} \implies 4+9=5$$ and this is definitely wrong.

Answer 3

Your step from $x^2+x=2$ to $x+x=\sqrt 2$ is not correct. First you cannot take the square root of just one term on one side of the equation. The proper expression would be $\sqrt{x^2+x}=\sqrt 2$ which is more complicated, not less, so probably a step in the wrong direction. Second, taking a square root introduces a sign ambiguity. If you had $x^2=4$ and took the square root you need to get $x=\pm 2$ because $(-2)^2=4$ as well as $2^2=4$.

Answer 4

Why are there two results? Well $(-a)^2 = (a)^2$ and though there is a convention that the principal (non-negative) value of the square root is used when we take the square root, when we operate by squaring a number, there are two numbers which give the same result - one is positive and the other negative.

So the template equation $$x^2=a^2$$has two solutions $x=\pm a$ unless we happen to have $a=0$ when there is one solution.

The template equation can also be written in the form $x^2-a^2=(x+a)(x-a)=0$.

To solve a quadratic we want to put it in template form. This is often easiest if we multiply the equation $ax^2+bx+c=0$ by $4a$ before doing any further work.

Here we multiply by $4$ to get $$4x^2+4x-8=0$$

To get template form we then "complete the square" (in the general case this gives a term $(2ax+b)^2$). Here we recognise $4x^2+4x$ as the first two terms of the square $$(2x+1)^2=4x^2+4x+1$$ so we see that $$(2x+1)^2=9=3^2$$

And we now have a template form and we can deduce $2x+1=\pm 3$ - the two possibilities coming from the two signs of the square roots we get from the template.

Notice how, in using the template, we needed to deal with both the terms involving $x$, so that we could take the square root - this is what we mean by "completing the square". Since we don't know what $x$ is, if we don't do this, we are left with terms we can't evaluate.

Answer 5

Here is a step by step method:

\begin{align} x^2 + x - 2 = 0 \\ x^2 + 2\frac{1}{2}x + \frac{1}{4} - \frac{1}{4} -2 = 0 &&&\text{Completing the square} \\ (x + \frac{1}{2})^2 - \frac{1}{4} - 2 = 0 &&& a^2+2ab+b^2 = (a+b)^2 \\ (x + \frac{1}{2})^2 - \sqrt{\frac{1}{4} + 2}^2 = 0 \\ \left(x+\frac{1}{2}-\sqrt{\frac{1}{4} + 2}\right)\left(x+\frac{1}{2}+\sqrt{\frac{1}{4} + 2}\right) = 0 &&& a^2-b^2 = (a-b)(a+b) \end{align}

We have found our solution $x^\pm = -\frac{1}{2}\pm\sqrt{\frac{1}{4}+2}$. These are exactly the same as the solutions of the form $\displaystyle x^\pm = -\frac{b\pm\sqrt{b²-4ac}}{2a}$ that you want. Note that $\sqrt{\frac{1}{4}+2} = \sqrt{\frac{9}{4}} = \frac{3}{2}$, so we can simplify: \begin{align} \left(x+\frac{1}{2}-\frac{3}{2}\right)\left(x+\frac{1}{2}+\frac{3}{2}\right) = 0 \\ \left(x-\frac{2}{2}\right)\left(x+\frac{4}{2}\right) = 0 \\ \left(x-1\right)\left(x+2\right) = 0 \end{align} Yielding the solutions $x_1=1$ and $x_2 = -2$.

Answer 6

$x^{2}+x=2$ does not imply $x+x=\sqrt{2}$

In fact $x^{2}+x$ does not have a simpler square-root, but $x^2+x+\frac14$ does because it is $\left(x+\frac12\right)^2$

So let's try that:

$$x^{2}+x-2=0$$ $$x^{2}+x+\frac14=\frac94$$ $$x+\frac12=\frac32 \text{ or } -\frac32$$ $$x=1 \text{ or } -2$$