Question

you can use a system of equations to graph and solve the polynomial equation $3x^3 + x = 2x^2 + 1$. which statement is true?
the $x$-coordinates of the solutions to the system and the zeroes of the equation are not equal.
the system has one solution, and the equation has three zeroes.
the equation has one zero, and the system has three solutions.
the $y$-coordinates of the solutions to the system and the zeroes of the equation are not equal.

Explanation:

Step1: Rewrite the polynomial equation

Rewrite \(3x^{3}+x = 2x^{2}+1\) as \(3x^{3}-2x^{2}+x - 1=0\). This is a cubic equation (degree 3), so by the Fundamental Theorem of Algebra, it has 3 roots (zeros), which can be real or complex. But when we solve it as a system of equations, we set \(y = 3x^{3}+x\) and \(y=2x^{2}+1\). The solutions to the system are the points of intersection of the two graphs, and the \(x\)-coordinates of these intersection points are the solutions to the original equation (the zeros of the equation \(3x^{3}-2x^{2}+x - 1 = 0\)).

Step2: Analyze the degree and number of solutions

The equation \(3x^{3}+x=2x^{2}+1\) is a cubic equation (degree 3), so it can have up to 3 real zeros (or a combination of real and complex, but for the purpose of graphing the system \(y = 3x^{3}+x\) and \(y = 2x^{2}+1\), we consider real solutions). The system of equations \(y=3x^{3}+x\) and \(y = 2x^{2}+1\) will have solutions where \(3x^{3}+x=2x^{2}+1\), so the \(x\)-coordinates of the system's solutions are the zeros of the original equation. Now, let's analyze the options:
- Option 1: Says \(x\)-coordinates of system solutions and zeros of equation are not equal. But they should be equal (since system solutions' \(x\)-coordinates satisfy the equation, so they are the zeros). So this is false.
- Option 2: Let's consider the graphs. The cubic function \(y = 3x^{3}+x\) has a general shape that goes from \(-\infty\) to \(+\infty\) (since leading coefficient positive), and the quadratic \(y=2x^{2}+1\) is a parabola opening upwards. A cubic and a quadratic can intersect at 1, 2, or 3 points? Wait, the cubic equation \(3x^{3}-2x^{2}+x - 1=0\). Let's check the number of real roots. We can use the Rational Root Theorem: possible rational roots are \(\pm1,\pm\frac{1}{3}\). Testing \(x = 1\): \(3 - 2+1 - 1=1 eq0\). \(x=\frac{1}{3}\): \(3(\frac{1}{27})-2(\frac{1}{9})+\frac{1}{3}-1=\frac{1}{9}-\frac{2}{9}+\frac{3}{9}-1=\frac{2}{9}-1=-\frac{7}{9} eq0\). \(x=-1\): \(-3 - 2 - 1 - 1=-7 eq0\). Now, let's check the derivative of \(f(x)=3x^{3}-2x^{2}+x - 1\), \(f^\prime(x)=9x^{2}-4x + 1\). The discriminant of \(f^\prime(x)\) is \(16 - 36=-20<0\), so \(f(x)\) is always increasing (since derivative is always positive). So \(f(x)\) has only one real root (because it's strictly increasing, so crosses the \(x\)-axis once). So the system \(y = 3x^{3}+x\) and \(y=2x^{2}+1\) has one solution (since \(f(x)\) is strictly increasing, so only one \(x\) where \(3x^{3}+x=2x^{2}+1\)), and the original equation (cubic) has three zeros? Wait, no, a cubic equation has three zeros (counting multiplicities, and complex zeros come in pairs). But if the derivative shows it's strictly increasing, then it has only one real zero and two complex conjugate zeros. Wait, but the option says "the system has one solution, and the equation has three zeroes". The equation (cubic) does have three zeroes (one real, two complex), and the system (which finds real solutions, since we graph in real plane) has one solution (the real zero). So this option is correct? Wait, let's re - check.
- Option 3: Says "the equation has one zero, and the system has three solutions". But the equation is cubic, so it has three zeros (counting complex), and the system (real solutions) has one solution (as we saw the function is strictly increasing, so only one real intersection). So this is false.
- Option 4: Talks about \(y\)-coordinates. The zeros of the equation are \(x\)-values (where \(y = 0\) for the equation \(3x^{3}+x-2x^{2}-1 = 0\)), but the \(y\)-coordinates of the system solutions are \(y=2x^{2}+1\) (or \(y = 3x^{3}+x\)) at the intersection points. This is a bit of a misdirection, and the key is about \(x\)-coordinates. So this option is not relevant.

Answer:

The system has one solution, and the equation has three zeroes. (The option with this statement)