Algebra 1 "Algebra I Reasoning with Equations and Inequalities" - Public Release - MSDE

Correct Answer

Option C (c is greater than 0) should be selected.

Quick sketch of f of x (blue) and g of x

Functions f and g plotted on the xy-plane.

Graphical Explanation

Since the y coordinate for both points of intersection is positive (the graphs intersect at two points above the x-axis), the value of c must be positive. That is sufficient to answer the question. A more detailed algebraic solution is provided in the "Algebraic Solution" tab.

Explanation

To determine which of the given inequalities must be true, we need to find where the point coordinates b, c lies on both the graph of f of x equals the opposite of x plus 2 and g of x equals x squared minus 1. This point coordinates b, c satisfies both functions simultaneously.

Since the point lies on both graphs, it must satisfy: c equals f of b and c equals g of b. This means: f of b equals g of b. Substituting the expressions for f of b and g of b we getthe opposite of x plus 2 equals x squared minus 1.

Now solve this quadratic equation using the quadratic formula:x equals fraction with numerator of the opposite of b plus or minus the square root of b squared minus 4 a c, end square root, and denominator of 2 a.

In this case, a equals 1, b equals 1, and c equals negative 3 x equals fraction with numerator of negative 1 plus or minus the square root of 1 squared minus 4 open parenthesis 1 close parenthesis open parenthhesis negative 3 close parenthesis, end square root, and denominator of 2 open parenthesis 1 close parenthesis.

Thus, the solutions for b are:b equals fraction with numerator of negative 1 plus the square root of 13 and denominator of 2. And. B equals fraction with numerator of negative 1 minus the square root of 13 and denominator of 2.

Since the square root of 13 is approximately equal to 3 point 6 these solutions are b is approximately equal to fraction with numerator of negative 1 plus 3 point 6 and denominator of 2, which equals fraction with numerator of 2 point 6 and denominator of 2, which equals 1.3. and b is approximately equal to fraction with numerator of negative 1 minus 3 point 6 and denominator of 2, which equals fraction with numerator of negative 4 point 6 and denominator of 2, which equals negative 2.3. Therefore, b has one positive value and one negative value.

To find corresponding c values, substitute both values of b into f of b equals the opposite of b plus 2. c is approximately equal to f of 1.3, which equals negative 1.3 plus 2, which equals 0 point 7. C is approximately equal to f of negative 2.3, which equals negative of negative 2.3 plus 2, which equals 2 point 3 plus 2, which equals 4 point 3.

So c is greater than 0 for both values of b.

Find the solutions by factoring

The given equation is:x squared plus 14 x minus 32 equals 0.

We could try to factor this quadratic expression into the form open parenthesis x plus a close parenthesis, open parenthesis x plus b close parenthesis, where a and b are numbers that satisfy:

1. a times b equals negative 32 (the constant term)
2. a plus b equals 14 (the coefficient of the middle term)

We need to find two numbers that multiply to negative 32 and add to 14. The pair that works is 16 and negative 2. Using a equals 16 and b equals negative 2, the equation can be written as:open parenthesis x plus 16 close parenthesis, open parenthesis x minus 2 close parenthesis, equals 0.

According to the zero-product property, if open parenthesis x plus 16 close parenthesis, open parenthesis x minus 2 close parenthesis, equals 0, then either x plus 16 equals 0 or x minus 2 equals 0 (or both). x plus 16 equals 0 gives x equals negative 16. x minus 2 equals 0 gives x equals 2.

Find the solutions using the quadratic formula

To find the two solutions to the equation you can also use the quadratic formula, although this particular equation is much easier solved by factoring. The quadratic formula is a standard method for solving quadratic equations:x equals fraction with numerator of the opposite of b plus or minus the square root of b squared minus 4 a c, end square root, and denominator of 2 a.

For a quadratic equation in the form a x squared plus b x plus c equals 0, a, b, and c are the coefficients. Here, a equals 1, b equals 14, and c equals negative 32. This gives usx equals fraction with numerator of negative 14 plus or minus the square root of 14 squared minus 4 open parenthesis 1 close parenthesis open parenthesis negative 32 close parenthesis, end square root, and denominator of 2 open parenthesis 1 close parenthesis.

This equation simplifies tox equals fraction with numerator of negative 14 plus or minus 18 and denominator of 2.

That gives us two solutions:x equals 2 and x equals negative 16.