Let sqrt(x)=t and t is positive. The given equation then reads t^4-38t^2 -t - 1656 = 0. t=8 is a solution. [t^4-38t^2 -t - 1656]/(t-8) = t^3 + 8 t^2 + 26t + 207. t^3 + 8 t^2 + 26t + 207=0 has one real solution which is negative and therefore unacceptable. Thus, t=8 > x = 64.
You haven’t shown that the cubic only has 1 real root even though I agree with you, it could be the case that all the roots are real. But because t cannot be negative regardless you got bailed out. If t could be negative then you would still have to show that the other two can be rejected because they are not real.
Synthetic division can be used to calculate function values. 1 0 -38 -1 -1656 (6) 6 36 -12 -78 6 6 -2 -13 -1734 By the way, 13:22 must be a^3+8a^2+26a+207=0.😎
For the cubic equation whose coefficients are only positive numbers then we may get the solutions either all negative or complex but surely no positive root is there. That's why the cubic equation won't be accepted. Thanks
The better argument is to state that there are two cases with the cubic equation, either this equation has 3 real roots which in this case would be all negative and can be rejected or you have 1 real root and 2 complex roots which can also be rejected since we are looking for reals and our restriction on x. Also I want to note that looking at sign changes doesn’t tell you that the solutions must be negative because we see repeated addition. Descartes Rule of sign talks about only negative or positive roots and doesn’t guarantee that they are positive or negative because you can have a function such that it doesn’t change sign at all for either positive or negative values and can yield complex solutions or you can get simply 0.
In all possible arguments one point is common that we won't get a single positive real root, so we can easily reject cubic equations for real solutions. Thanks
@@infyGyan that’s only because how you defined “a” to be sqrt(x) in general the cubic can have 3 negative real roots so it’s not good enough to say that the cubic has 1 negative and 2 complex roots.
Let sqrt(x)=t and t is positive. The given equation then reads t^4-38t^2 -t - 1656 = 0. t=8 is a solution. [t^4-38t^2 -t - 1656]/(t-8) = t^3 + 8 t^2 + 26t + 207. t^3 + 8 t^2 + 26t + 207=0 has one real solution which is negative and therefore unacceptable. Thus, t=8 > x = 64.
You haven’t shown that the cubic only has 1 real root even though I agree with you, it could be the case that all the roots are real. But because t cannot be negative regardless you got bailed out. If t could be negative then you would still have to show that the other two can be rejected because they are not real.
Nearby perfect square is 2025= 45^2
Diffrence is 8
Add √x= 8 i.e x= 64
64-45= 19
Synthetic division can be used to calculate function values. 1 0 -38 -1 -1656
(6) 6 36 -12 -78
6 6 -2 -13 -1734 By the way, 13:22 must be a^3+8a^2+26a+207=0.😎
Dear Sir How do you know there are two complex and one real solution @10:30 ? Thanks for your great lessons 🙂
For the cubic equation whose coefficients are only positive numbers then we may get the solutions either all negative or complex but surely no positive root is there. That's why the cubic equation won't be accepted.
Thanks
The better argument is to state that there are two cases with the cubic equation, either this equation has 3 real roots which in this case would be all negative and can be rejected or you have 1 real root and 2 complex roots which can also be rejected since we are looking for reals and our restriction on x. Also I want to note that looking at sign changes doesn’t tell you that the solutions must be negative because we see repeated addition. Descartes Rule of sign talks about only negative or positive roots and doesn’t guarantee that they are positive or negative because you can have a function such that it doesn’t change sign at all for either positive or negative values and can yield complex solutions or you can get simply 0.
In all possible arguments one point is common that we won't get a single positive real root, so we can easily reject cubic equations for real solutions.
Thanks
@@infyGyan that’s only because how you defined “a” to be sqrt(x) in general the cubic can have 3 negative real roots so it’s not good enough to say that the cubic has 1 negative and 2 complex roots.
Finally➡️ x= 64 .....thank you
🤗😊