(-x)cdot (x-43)=0 (3-x)cdot (x+4)=(1)

The first equation is already solved in the example. The solution is x = 0, x = 43.<br /><br />The second equation, $(3-x)(x+4) = 1$, is a quadratic equation and requires a different approach than the simple zero-product property used in the examples. We need to expand and rearrange it into standard form:<br /><br />(3-x)(x+4) = 1<br />3x + 12 - x² - 4x = 1<br />-x² - x + 12 = 1<br />-x² - x + 11 = 0<br />x² + x - 11 = 0<br /><br />This quadratic equation doesn't factor easily. We would need to use the quadratic formula to solve for x:<br /><br />x = [-b ± √(b² - 4ac)] / 2a<br /><br />where a = 1, b = 1, and c = -11.<br /><br />Therefore:<br /><br />x = [-1 ± √(1² - 4 * 1 * -11)] / (2 * 1)<br />x = [-1 ± √(1 + 44)] / 2<br />x = [-1 ± √45] / 2<br />x = [-1 ± 3√5] / 2<br /><br />So the two solutions are:<br /><br />x = (-1 + 3√5) / 2 and x = (-1 - 3√5) / 2<br /><br /><br />In summary:<br /><br />* **$(-x)(x-43) = 0$**: x = 0, x = 43<br />* **$(3-x)(x+4) = 1$**: x = (-1 + 3√5) / 2, x = (-1 - 3√5) / 2<br />