Question

If $x^2-2x-15=(x+r)(x+s)$ for all values of $x,$ and if $r$ and $s$ are constants, then which of the following is a possible value of $r-s?$

Short answer

Possible values of r-s are 8 and -8.

Step-by-step solution

Recognize the equation format

The given equation is in the quadratic form: $$x^2-2x-15=(x+r)(x+s)$$. The standard form of a quadratic equation is $$A{x}^2+Bx+C=0$$.

Expand the right side

Expand the right side of the equation: $$(x+r)(x+s)=x^2+(r+s)x+rs$$.

Equate coefficients

From the expanded form, we can equate coefficients with the given quadratic equation. This provides: 1. $$r+s=-2$$ (coefficient of x).2. $$rs=-15$$ (constant term).

Solve for r and s

Solve the system of equations: $$r+s=-2$$ $$rs=-15$$. To find the roots, substitute these into the quadratic formula: $$t^2+2t-15=0$$.

Use the quadratic formula

The quadratic formula is: $$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$. Here, $a=1$, $b=2$, and $c=-15$. Substitute these values in: $$t=\frac{-2\pm \sqrt{(2{)}^2-4\cdot 1\cdot (-15)}}{2\cdot 1}$$ $$t=\frac{-2\pm \sqrt{4+60}}{2}$$ $$t=\frac{-2\pm \sqrt{64}}{2}$$ $$t=\frac{-2\pm 8}{2}$$. This gives $t=3$ or $t=-5$. Therefore, $r$ and $s$ can be 3 and -5, respectively.

Calculate r-s

Using the possible values found: $r=3$ and $s=-5$, $r-s=3-(-5)=3+5=8$. Another possible set would be $r=-5$ and $s=3$, leading to $r-s=-5-3=-8$.

Key concepts

quadratic formula

The quadratic formula is crucial for solving equations of the form $a{x}^2+bx+c=0$.
It helps find the roots (solutions) by using the formula:
$$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
Let's break this down: the formula consists of three main parts,

• The numerator involves $\pm$ (plus or minus). This means we will have two solutions because we will add and subtract the square root value.
• The term inside the square root $\sqrt{b^2-4ac}$ is called the discriminant. The discriminant determines the nature of the roots.
• The denominator 2a provides the factor by which the entire result is divided.

By substituting the values of a, b, and c from our quadratic equation into the formula, we can solve for t. In the provided solution, $t=\frac{-2\pm 8}{2}$ gives us the roots 3 and -5.
Therefore, the quadratic formula simplifies the process of finding the values of unknown variables in quadratic equations.

equating coefficients

Equating coefficients is a method used to solve quadratic equations by matching coefficients from both sides of the equation.
Let's say you have an equation like this:
$x^2-2x-15=(x+r)(x+s)$
When you expand the right-hand side, you get:
$(x+r)(x+s)=x^2+(r+s)x+rs$
In this step, we match the coefficients of the corresponding terms from both sides of the equation.
By doing this, we get:

• $r+s=-2$ (coefficient of x)
• $rs=-15$ (constant term)

These equations provide relationships between r and s, helping in solving them accurately. In other words, it's just about comparing the terms on each side of the equation to find the values of unknown coefficients. This method allows us to break complex problems into manageable smaller parts.

roots of equations

The roots of quadratic equations are the values of x that satisfy the equation. Essentially, roots represent the points where the graph of the quadratic equation intersects the x-axis.
In the original problem, we started with:
$x^2-2x-15=0$
Expanding and solving by equating coefficients, we reach two equations:

• $r+s=-2$
• $rs=-15$

To find the values of r and s, we solved the quadratic equation derived from these, which was:
$t^2+2t-15=0$.
By applying the quadratic formula, we obtained the roots, t = 3 and t = -5.
These roots are the values that r and s can take. Hence, the roots reveal the essential solutions of the equation, crucial for understanding and resolving quadratic expressions.