The quadratic equation has two distinct real roots and such that . What is the value of the constant ?

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SAT Nonlinear Functions Question 131 | Free SAT Prep | Aniko

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Question 131·Medium·Nonlinear Functions

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The quadratic equation

x^2 - 5x + c = 0

has two distinct real roots $r$ and $s$ such that $r - s = 1$ . What is the value of the constant $c$ ?

For quadratic equations with conditions on the roots (like knowing their sum, product, or difference), quickly use the standard relationships: for $x^2+bx+c=0$ , the sum of the roots is $-b$ and the product is $c$ . Combine these with any extra information (such as a given difference) to set up a simple system of equations in the roots, solve for the roots themselves, and then use the product to get the unknown coefficient. This avoids writing the quadratic formula or testing answer choices and is both faster and less error-prone on the SAT.

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Connect roots to coefficients

For a quadratic of the form $x^2 - 5x + c = 0$ with roots $r$ and $s$ , think about how $r+s$ and $rs$ relate to the coefficients of $x$ and the constant term.

Use the given relationship between the roots

You know $r - s = 1$ and you can find $r + s$ from the coefficient of $x$ . How can you use these two equations together to solve for $r$ and $s$ ?

Relate c to r and s

Once you know the actual values of $r$ and $s$ , recall which expression involving $r$ and $s$ equals the constant term of the quadratic.

Desmos Guide

Graph the equations for r and s

Treat $r$ as $x$ and $s$ as $y$ . In Desmos, enter the two lines that represent the relationships between the roots:

Type y = x - 1 (this comes from $r - s = 1$ , or $s = r - 1$ ).
Type y = 5 - x (this comes from $r + s = 5$ , or $s = 5 - r$ ).

Find the intersection point (the roots)

Click on the point where the two lines intersect. Desmos will display its coordinates $(x,y)$ ; these values correspond to the roots $r$ and $s$ of the quadratic.

Use the intersection to find c

In a new expression line, multiply the two coordinates from the intersection point (the $x$ -value times the $y$ -value). This product equals $rs$ , which is the value of the constant $c$ in the quadratic.

Step-by-step Explanation

Relate the roots to the coefficients

For a monic quadratic with roots $r$ and $s$ ,

x^2 - (r+s)x + rs = 0

Comparing this to $x^2 - 5x + c = 0$ gives:

$r + s = 5$
$rs = c$

So we know the sum and product of the roots in terms of the coefficients.

Use the given difference of the roots

We are told that $r - s = 1$ .

Now we have a system of two equations:

\begin{aligned} r + s &= 5 \\ r - s &= 1 \end{aligned}

Add these equations to solve for $r$ :

(r+s) + (r-s) = 5 + 1 \Rightarrow 2r = 6 \Rightarrow r = 3.

Subtract the second equation from the first to solve for $s$ :

(r+s) - (r-s) = 5 - 1 \Rightarrow 2s = 4 \Rightarrow s = 2.

So the roots are $r=3$ and $s=2$ (in some order).

Use the product of the roots to find c

From Step 1, we know $rs = c$ .

Now substitute the values of $r$ and $s$ :

rs = 3 \cdot 2 = 6.

Therefore, $c = 6$ , so the correct answer choice is $6$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation