# What does b equal?

**Answer # 1 #**

The general shape of the graph of all quadratic functions is a parabola. The only exception is when the a is 0. Then the graph is a straight line, since we no longer have a quadratic whose highest power is 2, but a linear function whose highest power is 1.

Let's look at a random quadratic function to see what the graph looks like; then we will see how the b-value affects this graph. While changes in the a and c value also affect the graph, in this lesson we're focusing on how changes in the b-value alone affect the graph.

Let's look at the graph of f(x) = x^2 + 3x + 1, which is below. The b-value in this equation is 3.

We see that our graph is indeed a parabola. Our parabola is curving up. The x-value of the vertex, the tip of the parabola, is -3 / 2 or -1.5. We can actually calculate this x-value by evaluating the expression -b / 2a, where a and b are the values from the quadratic function. Our function has an a of 1 and a b of 3, so plugging these into the expression -b / 2a gives us -3 / 2 * 1 = -3 / 2 or -1.5, as expected. The point where the graph crosses the y-axis is given by our c-value. Our c is 1, and our graph crosses the y-axis at 1, as expected.

Now, what happens when we start changing the value of b? Let's see. We're going to keep our other values, a and c, constant while we play around with b to see what changes. Right now our a is positive, so let's see what happens to b when our a is positive.

Changing our b to 2, we get this kind of graph:

What has changed? It looks like our graph has shifted up and to the right. The x-value of our vertex is now at -1.

Okay, so our graph is shifting with the change in b; but what kind of overall shifting is occurring? Let's continue to play.

Let's change our b to 1.

Our vertex has moved to where x equals -1/2 or -0.5.

What about when b equals 0, -1, -2, and -3? Let's see:

Pretty interesting, isn't it? Our parabola continues to shift to the right as our b gets smaller and smaller. The vertex of our parabola also seems to be moving along a parabola of its own, with the tip happening when b is 0. Let's see how all the graphs look stacked on top of each other:

**Answer # 2 #**

We have seen that, in the case when a parabola crosses the \(x\)-axis, the \(x\)-coordinate of the vertex lies at the average of the intercepts. Thus, if a quadratic has two real roots \(\alpha, \beta\), then the \(x\)-coordinate of the vertex is \(\dfrac{1}{2}(\alpha+\beta)\). Now we also know that this quantity is equal to \(-\dfrac{b}{2a}\). Thus we can express the sum of the roots in terms of the coefficients \(a,b,c\) of the quadratic as \(\alpha+\beta = -\dfrac{b}{a}\).

In the case when the quadratic does not cross the \(x\)-axis, the corresponding quadratic equation \(ax^2+bx+c=0\) has no real roots, but it will have complex roots (involving the square root of negative numbers). The formula above, and other similar formulas shown below, still work in this case.

We can find simple formulas for the sum and product of the roots simply by expanding out. Thus, if \(\alpha, \beta\) are the roots of \(ax^2+bx+c=0\), then dividing by \(a\) we have

Comparing the first and last expressions we conclude that

From these formulas, we can also find the value of the sum of the squares of the roots of a quadratic without actually solving the quadratic.

Note that, in the previous exercise, the desired expressions are symmetric. That is, interchanging \(\alpha\) and \(\beta\) does not change the value of the expression. Such expressions are called symmetric functions of the roots.

You may recall that the arithmetic mean of two positive numbers \(\alpha\) and \(\beta\) is \(\dfrac{1}{2}(\alpha + \beta)\), while their geometric mean is \(\sqrt{\alpha\beta}\). Thus, if a quadratic has two positive real solutions, we can express their arithmetic and geometric mean using the above formulas.

**Answer # 3 #**

Now $$a^2= b^2 \iff (-a)^2 = a^2 = b^2 = (-b)^2 $$

And that's why, $$|a|=|b| \iff |a|=±b$$ but $$|a|=-b $$ is not possible as modulus of any thing cannot be negative. So $$|a| = b$$

$$ \iff ±a=b $$

$$ \text{OR} \ \ \ \ \ a=b , a=-b \ \ \ \ \ \square $$

I would like to give $ \text{ONE MORE} $ proof.

We have, $$|a|=|b|$$