Quadratic Roots Problems

[varun_kanpur]

I tried solving this problem but I couldn't

Given that a and b are the roots of equation x^2=7x+4
Show that a^3=53a+28

 

[stapel]

I tried solving this problem but I couldn't

What did you try? How far did you get? We can't help you find errors in work that we can't see. :shock:

Given that a and b are the roots of equation x^2=7x+4
Show that a^3=53a+28

I suspect that you're being expected to use the facts relating the sum and the product of the roots "a" and "b" to the coefficients in the quadratic. If you set up the relevant equations, solve for "b=" in each, and set those results equal to each other, you should be able to obtain a quadratic in "a" only. Isolate "a^2" and then multiply both sides by "a". In the result, substitute for the "a^2", and simplify. You should get the desired result.

 

[Deleted member 4993]

I tried solving this problem but I couldn't

Given that a and b are the roots of equation x^2=7x+4
Show that a^3=53a+28

your equation is x² - 7*x - 4 = 0

The roots of this equation are:

a = ??

b = ??

now evaluate a³ - 53*a -28 = ??

 

[JeffM]

I tried solving this problem but I couldn't

Given that a and b are the roots of equation x^2=7x+4
Show that a^3=53a+28

One way to solve this question is to find a rational root of the function \(\displaystyle y^3 - 53y - 28.\) (There is in fact an integer root.) EDIT: I should point out that this is not the fastest way to solve this problem, and finding a rational root is just the first step. I put it first because it was the way that I first solved the problem myself. For some reason, it was an intuitive approach for me.

There are two other ways to solve this problem, one is quite quick and little prone to error, and the other is lengthy and very prone to error. Because you have not shown your work, we cannot tell where you have made an error. Not only would it help a lot if you showed your work, it would also help if you were to explain what you are studying.

 

[HallsofIvy]

As JeffM suggested, actually solving the equation is not a good way to go. You are told that \(\displaystyle a^2- 7a- 4= 0\) so that \(\displaystyle a^2= 7a+ 4\). So \(\displaystyle a^3= a(a^2)= a(7a+ 4)= 7a^2+ 4a\). Now use \(\displaystyle a^2= 7a+ 4\) again.

 

[Deleted member 4993]

As JeffM suggested, actually solving the equation is not a good way to go. You are told that \(\displaystyle a^2- 7a- 4= 0\) so that \(\displaystyle a^2= 7a+ 4\). So \(\displaystyle a^3= a(a^2)= a(7a+ 4)= 7a^2+ 4a\). Now use \(\displaystyle a^2= 7a+ 4\) again.

However, in these days of calculators with storage buttons - actually solving the equation and evaluating the second equation would be fastest (less elegant though).