Step by step
The question
Question: for which k does kx² + 4x + 1 = 0 have exactly one solution?
1 Find a, b and c
We read the numbers.
With x² is the letter k.
b = 4 is with x.
c = 1 is the plain number.
2 Choose the rule
From the discriminant lesson: the sign of b² − 4ac tells how many solutions there are.
We want exactly one solution.
So b² − 4ac must be 0.
We write it: b² − 4ac = 0.
3 Put the numbers in
We put in the numbers.
The letter k stays as it is: it is the number we want.
Careful!
4² means 4 times 4, not times 2.
4 squared is 16.
4 times k times 1 is 4k.
Now we have 16 − 4k = 0.
We want k by itself.
Two things are stuck to k: times −4 and + 16.
First k was times −4.
Then + 16.
To take them off, we go backwards.
First we remove the + 16.
Then the times −4.
4 Remove the 16
To remove 16, we take away 16.
We do it on both sides, so they stay equal.
16 and − 16 make zero.
Cross them out.
0 − 16 = −16.
Now we have −4k = −16.
5 Remove the −4
−4k means −4 times k.
We want just k.
To remove times −4, we divide by −4.
We do it on both sides, so they stay equal.
−4 divided by −4 is 1.
So just k is left.
−16 divided by −4 is 4.
So k = 4!
6 Check
Let's check.
We put k = 4 back in.
b² − 4ac is 16 take away 16: 0.
In the quadratic formula, the square root of 0 is 0.
Plus 0 and minus 0 give the same number: x = −1/2.
Exactly one solution.
It works!
Let's look back.
First: find a, b and c.
Then: choose the rule.
Then: put the numbers in.
Then: remove the 16.
Then: remove the −4.
Last: check.