Question

The function \(g\) is defined by \(g(x)=2 x^2-d x-6\), where \(d\) is a constant. If one of the zeros of \(g\) is 6 , what is the value of the other zero of \(g\) ? A) 2 B) \(\frac{1}{2}\) C) \(-\frac{1}{2}\) D) \(-2\)

Short answer

The other zero of $g(x)$ is \(-\frac{1}{2}\), so the answer is (C).

Step-by-step solution

Determine the value of d

Since 6 is a zero of the function g(x), we substitute 6 into the function and set it equal to 0: \(g(6)=2(6)^2 - d(6) - 6 = 0\) Solve for d by simplifying the equation: \(72-6d-6=0\) \(66-6d=0\) Now, divide both sides by 6: \(11-d=0\) Thus, d=11.

Substitute the value of d into the function

Now we know the value of d is 11, we can substitute it into the function g(x): \(g(x)=2x^2-11x-6\)

Factor the quadratic equation

We need to factor the quadratic equation to find the other root (zero): \(2x^2-11x-6 = (2x+a)(x+b)\) The factors of 2x² are (\(2x\), \(x\)). The factors of -6 are (\(-1, 6\), \(1, -6\), \(-2, 3\), \(2, -3\)). By trying different combinations of the numbers, we find that the correct factorization of the quadratic equation is: \((2x+1)(x-6)\)

Solve for the other zero

We now have the factored equation which is \((2x+1)(x-6) = 0\) we can solve for x by setting each factor equal to zero: Case 1: \(2x+1=0\) \(2x=-1\) \(x=-\frac{1}{2}\) Case 2: (which we already know) \(x-6=0\) \(x=6\) The other zero of g(x) is \(-\frac{1}{2}\), so the answer is (C).

Key concepts

Quadratic Equations

A quadratic equation is a polynomial equation of the second degree, which means the highest exponent of the variable, usually denoted by x, is 2. The standard form of a quadratic equation is \(ax^2 + bx + c = 0\), where a, b, and c are coefficients and a ≠ 0. The solutions to a quadratic equation are known as the roots or zeros of the equation. These solutions can be found using various methods, including factoring, the quadratic formula, or completing the square.

In our exercise, we encountered a quadratic equation when solving for the zeros of the function \(g(x)\). By substituting one known zero (6) into the equation and solving for \(d\), we transformed the equation into a more 'standard' quadratic form, which then could be factored to find the remaining zero. It's essential to understand how to manipulate these equations algebraically to uncover the values of \(x\) that satisfy the equation, which in this scenario, led us to find the value of the other zero.

Factoring Polynomials

To factor polynomials is to break them down into the product of their simplest polynomial factors, much like factoring numbers is to express them as a product of prime numbers. Factoring is a powerful tool for simplifying problems involving polynomials, such as finding the zeros of a quadratic function.

When factoring the given quadratic equation in our example, \(2x^2-11x-6\), we looked for two numbers that multiply to give the product of the coefficient of \(x^2\) and the constant term (here, 2 and -6) and at the same time, add up to give the coefficient of \(x\) (here, -11). Through trial and error or systematically testing possibilities, the equation was successfully factored into \(2x+1\) and \(x-6\), from which we could easily find the zeros. This step is crucial and often requires practice to identify the right pairs that will factor the quadratic properly.

Solving Quadratic Zeros

The term 'solving quadratic zeros' refers to finding the values of \(x\) that make the quadratic equation equal to zero. These values are also called the roots or solutions of the equation. Once a quadratic equation is factored, as we did with \(2x^2-11x-6\), solving for the zeros becomes a more manageable task.

Each factor set to zero gives us a potential solution to the equation. So from the factored form \(2x+1)(x-6)=0\), we establish two cases: For \(2x+1=0\), solving for \(x\) gives us \(x=-\frac{1}{2}\) and for \(x-6=0\), we find \(x=6\), which was already given. It's important to note that every quadratic equation will have two solutions, which might be real or complex, depending on the discriminant \(b^2-4ac\). However, in the context of the SAT math practice, we typically focus on real solutions, as seen in this exercise, where both found zeros of the quadratic function were real numbers.