Solution by half interval search: Replace m by x. x³ +4x² - 3 = 0. Let y = f(x) = x³ +4x² - 3. We know that y will be negative infinity for x = negative infinity and y will be positive infinity for x = positive infinity. Therefore, because f(x) is continuous, there is at least one zero crossing. So, we can write a computer program which performs a half interval search. We start with a value of x which produces a negative y, let's call that lower value, and another value of x which produces a positive value of y, call that upper value, and compute the value of y at the midpoint. If y is positive, it becomes the new upper value, and, if y is negative, it becomes the new lower value. We compute the midpoint and determine a new upper or lower value. We stop when y is exactly or when y is very small. Let's designate the variable a as the value of x found. Now we factor (x - a) out of x³ +4x² - 3 and are left with a quadratic equation. We solve that equation using the quadratic formula, so the quadratic factors into (x - b) and (x - c). We now have 3 factors. For the given equation, one of the factors will be very close to x +1. We try x = -1 and find that it is an exact solution to x³ +4x² - 3 = 0. We factor out (x + 1) and are left with x² + 3x -3 = 0, which we solve with the quadratic formula and have 3 solutions total. Recall that k = x/4, so we must divide each solution by 4 to get the value of k for each solution. In the general case, the function y = f(x) will have either 1 zero crossing or 3. If there is only one zero crossing, the half interval search finds it. The other two solutions are imaginary numbers. If there are 3 zero crossings, the half interval search will not find the middle one unless a search point lands exactly on or within our search tolerance of it. Initially, if there are 3 zero crossings, all 3 are in the search range. At some point, the midpoint will find 2 zero crossings to one side and 1 to the other side, and it will then focus in on the single zero crossing. In this case, we do find an x value close to -1 by calculations and confirm x = -1 is exact. It is possible that x will have a radical all real solutions. So, we have 3 solutions with long strings of digits following the decimal point. It will take some creativity to find a radical expression which is exact for at least 1 of the 3. However, it will always be possible to prove that the exact value is, indeed, exact.
its oral question for me
Solution by half interval search: Replace m by x. x³ +4x² - 3 = 0. Let y = f(x) = x³ +4x² - 3. We know that y will be negative infinity for x = negative infinity and y will be positive infinity for x = positive infinity. Therefore, because f(x) is continuous, there is at least one zero crossing. So, we can write a computer program which performs a half interval search. We start with a value of x which produces a negative y, let's call that lower value, and another value of x which produces a positive value of y, call that upper value, and compute the value of y at the midpoint. If y is positive, it becomes the new upper value, and, if y is negative, it becomes the new lower value. We compute the midpoint and determine a new upper or lower value. We stop when y is exactly or when y is very small. Let's designate the variable a as the value of x found. Now we factor (x - a) out of x³ +4x² - 3 and are left with a quadratic equation. We solve that equation using the quadratic formula, so the quadratic factors into (x - b) and (x - c). We now have 3 factors. For the given equation, one of the factors will be very close to x +1. We try x = -1 and find that it is an exact solution to x³ +4x² - 3 = 0. We factor out (x + 1) and are left with x² + 3x -3 = 0, which we solve with the quadratic formula and have 3 solutions total. Recall that k = x/4, so we must divide each solution by 4 to get the value of k for each solution.
In the general case, the function y = f(x) will have either 1 zero crossing or 3. If there is only one zero crossing, the half interval search finds it. The other two solutions are imaginary numbers. If there are 3 zero crossings, the half interval search will not find the middle one unless a search point lands exactly on or within our search tolerance of it. Initially, if there are 3 zero crossings, all 3 are in the search range. At some point, the midpoint will find 2 zero crossings to one side and 1 to the other side, and it will then focus in on the single zero crossing.
In this case, we do find an x value close to -1 by calculations and confirm x = -1 is exact. It is possible that x will have a radical all real solutions. So, we have 3 solutions with long strings of digits following the decimal point. It will take some creativity to find a radical expression which is exact for at least 1 of the 3. However, it will always be possible to prove that the exact value is, indeed, exact.