Question

c) 4 - 2x - 3x² = 0
a = 4, b = -2, c = -3

Explanation:

Step1: Identify the quadratic equation form

The equation is \(4 - 2x - 3x^2 = 0\). Rewrite it in standard quadratic form \(ax^2+bx + c = 0\) by rearranging terms: \(-3x^2-2x + 4 = 0\) or multiply by -1: \(3x^2 + 2x - 4 = 0\) (Wait, but the given \(a = 4\), \(b=-2\), \(c = -3\) suggests the original equation was maybe miswritten? Wait, maybe the original equation is \(4x^2-2x - 3 = 0\)? Because if \(a = 4\), \(b=-2\), \(c=-3\), then the equation is \(4x^2-2x - 3 = 0\). Let's assume that's the case (maybe a typo in the handwritten note, swapping the coefficient of \(x^2\) and the constant term).

Step2: Use quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For \(a = 4\), \(b=-2\), \(c=-3\):

First, calculate the discriminant \(D=b^2 - 4ac=(-2)^2-4\times4\times(-3)=4 + 48 = 52\).

Then, \(x=\frac{-(-2)\pm\sqrt{52}}{2\times4}=\frac{2\pm2\sqrt{13}}{8}=\frac{1\pm\sqrt{13}}{4}\).

Wait, but maybe the original equation was \(4 - 2x - 3x^2 = 0\) (i.e., \(-3x^2-2x + 4 = 0\)), then \(a=-3\), \(b=-2\), \(c = 4\). But the given \(a = 4\), \(b=-2\), \(c=-3\) implies the equation is \(4x^2-2x - 3 = 0\). Let's proceed with \(a = 4\), \(b=-2\), \(c=-3\).

Discriminant \(D=(-2)^2-4(4)(-3)=4 + 48 = 52\).

Square root of 52 is \(2\sqrt{13}\), so:

\(x=\frac{2\pm2\sqrt{13}}{8}=\frac{1\pm\sqrt{13}}{4}\).

Answer:

The solutions are \(x=\frac{1+\sqrt{13}}{4}\) and \(x=\frac{1-\sqrt{13}}{4}\) (or if the equation was \(4 - 2x - 3x^2 = 0\), the solutions would be \(x=\frac{-2\pm\sqrt{(-2)^2-4(-3)(4)}}{2(-3)}=\frac{-2\pm\sqrt{4 + 48}}{-6}=\frac{-2\pm2\sqrt{13}}{-6}=\frac{1\mp\sqrt{13}}{3}\), but based on the given \(a,b,c\), the first calculation is correct).