help please?

Question

For what values of k has the following quadratic equation just one solution? 5x^2 - 6kx + k + 4 = 0


^= to the power of

it would be great if you could include the working out as this will help me understand it better

sorry i just saw that i wrote it wrong

5kx^2 - 6kx + k + 4 = 0

Answer 1

This is my all time favorite maths problem-it's brilliant:a quadratic inside of a quadratic!

Okay Here's my solution. Usually when finding a solution to a quad, I use the quadratic equation:

X=(-b +- ROOT(B^2 - 4ac)) / 2a

(Where a=the number before x^2, b=the number before x, and C=the constant).

This can give you your 2 values of x with plus or minus everything inside of the root. For there to be just the one solution, the square root of B^2 -4ac must equal 0. That way its

x= B+-0 / (2a)
Which will only give you 1 solution(since anything plus or minus 0 stays as it is).

So if you use the equation "B^2 -4ac=0", entering your values of A B and C you can find your values of k.


B^2 - 4ac=0

A=5
B=6k
C=K+4
Enter these values into the equation:
(6*k)^2 -4* (5)* (k+4)=0

Re-arrange:
36k^2 -20(k+4)=0
36k^2 -20k -80=0
Find the value of k using the normal quadratic equation:K=1.79414934 or K=-1.238593785

That's kinda right: except you will need to find these values exactly and express as surds-not decimals... I dont know how to do surds(hopefully you do). Then to double check enter this surd value of K into the original equation, and your should only find one value of x rather than the standard 2.

Answer 2

Ok... if you haven't learned the quadratic equation, then this probably won't make any sense to you. But if you have then you would know that x has two answers because of the positive and negative squareroot of ( b^2 - 4ac ). If you want to get only one answer for x, then you must get the negative and positive squareroots to be the same... That condition only exists at zero. So, to make the squareroot equal zero, Solve b^2 - 4ac = 0. And for the sake of simplicity, let c = k + 4. Solve for c. And then to solve for k, just subtract 4 from the answer. So, 5x^2 - 6kx + k + 4 = 0. a = 5, b = (-6k), and c = k + 4. Plug it in the b^2 - 4ac = 0 equation. You get 36k^2 - 4(5)(k+4) = 0 or 36k^2 - 20k - 80= 0. To make it easier, you could take out the Greatest Common Denomintor (or not, your choice). 4(9k^2 - 5k - 20) = 0. ( And then if you did take out the G.C.D., then you could divide both sides by 4. 9k^2 - 5k - 20 = 0. Now you have to solve for k. Plug this equation into the quadratic equation solving for k. You end up with (5 + sqrt(114)) / 18 and (5 - sqrt(114)) / 18.

Answer 3

56

Answer 4

I ended up with (x+1)(5x-6k)+(k-4)=0
since I couldnt factor out of the 5x-6k I cant give you a numerical answer
are you sure you typed the problem right?

Answer 5

sorry but i wish i could i could help you but i wish more that i could understand that

Answer 6

2p