Solve this its urgent

Question

If the sum of zeroes of p(x)=x^2-8x+k is 40, the find out k?

 

Its urgent answer this soon plzzz

Answer 1

If we're able to factor x² - 8x + k we will get something like (x - a)(x - b).

 

We factor quadratics of this form by asking ourselves what multiplies to give us k but adds to give us -8. -a and -b are what we get when we do this. 

 

This means that (-a)(-b) = ab = k and -a + (-b) = -8

which is the same as saying a + b = 8.

 

From the statement of the problem we have that the zeroes add to give us 40. The zeros are the numbers that make that factored expression above equal zero when we plug them in for x. Only a and b can do this (whatever they are).

 

This means that a + b = 40.

 

Well, hang on. When is it possible to have a + b = 8 AND a + b = 40? Never...

 

So maybe there's something wrong with my reasoning...maybe this is a question that isn't consistent with itself.

 

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Let's look at the solutions we get using the quadratic formula:

 

When x² - 8x + k = 0

 

our solutions will be

 

(1/2)[8 +- √(64 - 4k)]

 

(1/2)[8 +- 2√(16 - k)]

 

4 +- √(16 - k)

 

So our two solutions have to be

 

4 + √(16 - k)    and 

 

4 - √(16 - k)

 

But if we add these together, we get - still - 8.

 

It doesn't even matter what k is. The problem states that the solutions add to be 40, but they add to be 8 when I do both things I can think to do...so no wonder you're having trouble with this problem.

 

Still won't rule out that I'm babbling nonsense, but I think there might be something off about the statement of the problem.