Or you could use the method of bisection.
[MATH]y = x^2 ln(2x + 10) - (x - 6)ln(x^2 + x + 8) \implies x > - 5.[/MATH]
[MATH]\lim_{x \rightarrow -5} y = - \infty < 0.[/MATH]
[MATH]x = - 4 \implies y = 16ln(18) - (-12)ln(20) > 0.[/MATH]
Therefore y = 0 somewhere between - 5 and - 4. Try - 4.5. Keep on going until you have an approximate answer that is good enough. I trust Subhotosh so I am confident that you could get a better answer by starting with -4.69 and -4.67. How many decimal digits do you need?
Newton’s method converges a lot faster than bisection, but it requires some ugly looking calculus.
[MATH]y = x^2 ln(2x + 10) - (x - 6)ln(x^2 + x + 8) \implies x > - 5.[/MATH]
[MATH]\lim_{x \rightarrow -5} y = - \infty < 0.[/MATH]
[MATH]x = - 4 \implies y = 16ln(18) - (-12)ln(20) > 0.[/MATH]
Therefore y = 0 somewhere between - 5 and - 4. Try - 4.5. Keep on going until you have an approximate answer that is good enough. I trust Subhotosh so I am confident that you could get a better answer by starting with -4.69 and -4.67. How many decimal digits do you need?
Newton’s method converges a lot faster than bisection, but it requires some ugly looking calculus.