sat suite question viewer

Advanced Math / Nonlinear equations in one variable and systems of equations in two variables Difficulty: Hard

y = 2 x 2 - 21 x + 64

y = 3 x + a

In the given system of equations, a is a constant. The graphs of the equations in the given system intersect at exactly one point, (x,y), in the xy-plane. What is the value of x ?

Back question 65 of 148 Next

Explanation

Choice C is correct. It's given that the graphs of the equations in the given system intersect at exactly one point, x,y, in the xy-plane. Therefore, x,y is the only solution to the given system of equations. The given system of equations can be solved by subtracting the second equation, y=3x+a, from the first equation, y=2x2-21x+64. This yields y-y=2x2-21x+64-3x+a, or 0=2x2-24x+64-a. Since the given system has only one solution, this equation has only one solution. A quadratic equation in the form rx2+sx+t=0, where r , s , and t are constants, has one solution if and only if the discriminant, s2-4rt, is equal to zero. Substituting 2 for r , -24 for s , and - a + 64 for t in the expression s2-4rt yields -242-4264-a. Setting this expression equal to zero yields -242-4264-a=0, or 8 a + 64 = 0 . Subtracting 64 from both sides of this equation yields 8a=-64. Dividing both sides of this equation by 8 yields a=-8. Substituting -8 for a in the equation 0=2x2-24x+64-a yields 0=2x2-24x+64+8, or 0=2x2-24x+72. Factoring 2 from the right-hand side of this equation yields 0=2x2-12x+36. Dividing both sides of this equation by 2 yields 0=x2-12x+36, which is equivalent to 0=x-6x-6, or 0=x-62. Taking the square root of both sides of this equation yields 0 = x - 6 . Adding 6 to both sides of this equation yields x = 6

Choice A is incorrect. This is the value of a , not x .

Choice B is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.