In the system of equations above, is a constant. The graphs intersect at two points and , where and both -coordinates are positive. If , what is the value of ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 78 | Free SAT Prep | Aniko

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Question 78·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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y = x^2 + bx + 3

y = 5x - 3

In the system of equations above, $b$ is a constant. The graphs intersect at two points $(x_1, y_1)$ and $(x_2, y_2)$ , where $x_1 < x_2$ and both $x$ -coordinates are positive. If $x_2 - x_1 = 5$ , what is the value of $b$ ?

For intersection questions involving a line and a parabola, quickly set the expressions equal to form a quadratic in x; the roots of this quadratic are the x-coordinates of the intersection points. Use Vieta’s formulas (sum and product of roots) along with any extra information given (like the difference of the roots) to write equations in terms of the coefficients. Identities such as $(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2$ let you tie together sum, product, and difference efficiently, and then you can solve for the unknown parameter and check any extra conditions (like positivity) to select the valid solution.

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Form a single equation in x

At an intersection point, both equations have the same y-value. What equation do you get if you set $x^2 + bx + 3$ equal to $5x - 3$ and move all terms to one side?

Use what you know about roots of a quadratic

Once you have your quadratic in the form $x^2 + px + q = 0$ , recall that the sum of the roots is $-p$ and the product is $q$ . How can you express $x_1 + x_2$ and $x_1 x_2$ in terms of $b$ ?

Connect the given difference to the sum and product

You know $x_2 - x_1 = 5$ , and you can write $x_1 + x_2$ and $x_1 x_2$ in terms of $b$ . Try using the identity $(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2$ to get an equation involving only $b$ .

Check both algebraic possibilities

Solving your equation for $b$ should give two possible values. Use each to find the corresponding roots of the quadratic and check which choice gives two positive x-values that are 5 units apart.

Desmos Guide

Enter the general equations and understand the role of b

Type y = x^2 + b*x + 3 and y = 5x - 3. Desmos will treat b as a slider. The intersection x-values of these two graphs correspond to $x_1$ and $x_2$ for that b.

Test each answer choice for b

Use the slider or click on b and type each answer choice value (−7, −2, 8, 12) one at a time. For each b, click the intersection points that appear between the parabola and the line and note the x-coordinates.

Check the conditions for the correct b

For each tested value of b, verify two things from the intersection coordinates shown by Desmos: (1) both x-values are positive, and (2) the difference between the larger and smaller x-value is 5. The value of b that satisfies both conditions is the correct answer.

Step-by-step Explanation

Set the equations equal to find a quadratic for the intersection x-values

At intersection points, the y-values are equal, so set the right sides equal:

x^2 + bx + 3 = 5x - 3

Move everything to one side:

x^2 + bx + 3 - 5x + 3 = 0 \\[0.7em] x^2 + (b - 5)x + 6 = 0

The solutions of this quadratic are $x_1$ and $x_2$ , the x-coordinates of the intersection points.

Express the sum and product of the roots in terms of b

For a quadratic $x^2 + px + q = 0$ with roots $r_1$ and $r_2$ :

$r_1 + r_2 = -p$
$r_1 r_2 = q$

Here, $p = b - 5$ and $q = 6$ , and the roots are $x_1$ and $x_2$ . So:

x_1 + x_2 = -(b - 5) = 5 - b \\[0.7em] x_1 x_2 = 6

We are also given that $x_2 - x_1 = 5$ and both roots are positive.

Relate the difference of the roots to their sum and product

Use the identity that connects the sum, product, and difference of two numbers:

(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2

We know $x_2 - x_1 = 5$ , $x_1 + x_2 = 5 - b$ , and $x_1 x_2 = 6$ . Substitute these into the identity:

5^2 = (5 - b)^2 - 4(6)

25 = (5 - b)^2 - 24 \\[0.7em] (5 - b)^2 = 49

Now solve this equation for $b$ .

Solve for b and enforce the positivity condition

From $(5 - b)^2 = 49$ :

5 - b = 7 \quad \text{or} \quad 5 - b = -7

If $5 - b = 7$ , then $b = -2$ .
If $5 - b = -7$ , then $b = 12$ .

Now check which value gives positive roots for $x$ :

For $b = -2$ , the quadratic is $x^2 - 7x + 6 = 0$ , which factors as $(x - 1)(x - 6) = 0$ , so $x_1 = 1$ and $x_2 = 6$ . Both are positive and $x_2 - x_1 = 5$ .
For $b = 12$ , the quadratic is $x^2 + 7x + 6 = 0$ , which factors as $(x + 1)(x + 6) = 0$ , so $x_1 = -6$ and $x_2 = -1$ , which are not positive.

Therefore, the only value of $b$ that satisfies all conditions is $b = -2$ .

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation