One solution of kx^2 +6x-k=0 is -3. Find the other solution
I have the answer as -1/3 is this right?
The solution must depend upon k.
It is [-6 +/-sqrt(36 + 4k^2)]/2k
There might be a value of k that would make one root equal to -3, but I don't see it.
The product of two roots would have to be 1/(4k^2) * [36 - (36 + k^2)]
= -1/4
I do not agree with your answer.
How about doing this :
since x=-3
9k -18 -k = 0
k = 9/4
so your equation is (9/4)x^2 + 6x - 9/4 = 0
or 9x^2 + 24x - 9 = 0
3x^2 + 8x - 3 = 0
since one root is -3, you know one of the factors to be (x+3), the other can be easily reasoned out
(x+3)(3x-1) = 0
so x = -3, the given one or
x = 1/3
You were close.
To find the other solution of the quadratic equation kx^2 + 6x - k = 0, where one solution is -3, we can use the factored form of the quadratic equation:
kx^2 + 6x - k = (x - r1)(x - r2),
where r1 and r2 are the solutions of the equation. Since we know that -3 is a solution, we can substitute it into the factored form:
k(-3)^2 + 6(-3) - k = (-3 - r1)(-3 - r2).
Simplifying the equation, we get:
9k - 18 - k = (-3 - r1)(-3 - r2).
Combine like terms:
8k - 18 = (-3 - r1)(-3 - r2).
Now, we can use the fact that the product of the solutions of a quadratic equation is equal to the constant term divided by the coefficient of the quadratic term:
k = -18/8.
k = -9/4.
Substituting this value back into the equation:
8(-9/4) - 18 = (-3 - r1)(-3 - r2).
-18 - 18 = (-3 - r1)(-3 - r2).
-36 = (3 + r1)(3 + r2).
Now, we need to find two numbers, r1 and r2, whose product is -36 and whose sum is -6 (from the coefficient of the linear term in the original equation). We can solve this by trial and error or by using the quadratic formula.
By observing the factors (-4) and 9, we can see that (-4)(9) = -36 and (-4) + 9 = 5.
Therefore, the other solution is:
r1 = -4, r2 = 9.
So, the other solution of the equation kx^2 + 6x - k = 0, where one solution is -3, is -4.