SOLUTION: One solution of kx^2-5x+k=0 is 3. Find the other solution

Question 535981: One solution of kx^2-5x+k=0 is 3. Find the other solution
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Problem: if one solution of the quadratic equation:
.

.
is 3, what is the other solution?
.
Notice that when you term-by-term compare the given equation to the standard form of the quadratic equation:
.

.
you can see that "a" correlates to k, "b" equals -5, and "c" also equals k.
.
We know that the quadratic formula applies to solving a quadratic equation in the standard form. The quadratic formula says that for a quadratic equation in the standard form x can be found from:
.

.
into this formula we can substitute k for "a" and "c" and -5 for "b" to get:
.

.
The becomes + 5 and the and . These actions result in:
.

.
multiply both sides by 2*k which is the denominator on the right side. This clears the denominator, and the formula becomes:
.

.
we also know that one value for x is 3. We were told that in the statement of the problem. So substitute 3 for x in the formula. This makes the left side become:
.

.
multiply out the left side to get:
.

.
get rid of the 5 on the right side by subtracting 5 from both sides:
.

.
square both sides to get rid of the radical on the right side and this formula becomes:
.

.
cancel 25 on both sides of the formula by subtracting 25 from both sides. Also get rid of the on the right side by adding to both sides to get:
.

.
Factor a k from terms on the left side:
.

.
notice that this equation will be true if either of the two factors on the left side is equal to zero because multiplication by a zero results in the left side becoming zero and making the left side equal to the zero on the right side.
.
So this formula will be true if either k = 0 or if 40k - 60 = 0. We can ignore k = 0 because if this were true in the original quadratic equation of the problem if k were zero, both the
.
then dividing both sides by 60:
.

.
and reducing the right side by dividing both the numerator and denominator by 20 to get:
.

.
Now we can go back to the original problem and substitute for k to make it become:
.

.
Multiply all terms by 2 to cancel out the denominator and you have:
.

.
compared to the standard quadratic equation we now have "a" = 3, "b" = -10, and "c" = 3. Substituting these values into the quadratic formula, we get:
.

.
which becomes:
.

.
Simplifying to:
.

.
The radical term becomes the square root of 64:
.

.
which simplifies to:
.
and this simplifies to
.
and
.
and this simplifies to
.
The first value for x we already knew because it was part of the problem. The second value for x, which we found was , is the answer you were asked to find in the problem.
.
Hope this helps you to understand the quadratic equation a little better and how to apply that knowledge to solving this problem.
.
Check my work. It's late and I might have introduced some errors due to lack of coffee. You can also check this problem by multiplying the two factors:
.
and see if the answer does not give you the original equation in the problem in which
.
Good luck ...
.

RELATED QUESTIONS
One solution of {{{kx^2}}}-7x+k=0 is 4. Find the other... (answered by jim_thompson5910)

One solution of {{{ kx^2+3x-k=0 }}} is -2. Find the other solution using quadratic... (answered by rothauserc)

Find the value of "k" in this equation, if one solution is 1/2 x^2 - kx + 15 = 0 (answered by josgarithmetic)

If x=3,is a solution of the equation X^2+kx+15, find the value of k .Hence find the... (answered by josgarithmetic,ikleyn)

One solution of 2x^2+11x+k=0 is -5. Find the other... (answered by ankor@dixie-net.com)

One solution of 2x^2+11x+k=0 is -5. Find the other... (answered by checkley75)

Find the range of k for which the quadratic equation x^2-kx+3k-2=0 has two different... (answered by KMST)

Please help me solve this problem: Find the value of k. {{{6x^2+kx-6=0}}} ; one root is (answered by Edwin McCravy)

Find the number k such that the given equation has exactly one real solution:... (answered by Earlsdon)