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For the quadratic equation \(x^2\)+Kx+5; sum of roots = K and product of roots = 5

Roots given as (x+1) and (x+C)
Hence, product of the roots = 1*C = 5
Hence C=5

Sum of the roots = K = 1+5 = 6
Answer C
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Let \(C\) and \(K\) be constants.If \(x^2+Kx+5\) factors into \((x+1)*(x+C)\), then what is the value of \(K\)?

A)0
B)5
C)6
D)8
E)Cannot be determined

Source => NOVA.
Because we are given one factor's integer, (x+1), I would just factor \(x^2+Kx+5\).

With (x+1) and (x + ___), thinking about FOIL where last two terms multiplied must equal 5, we need:

1*__ = 5. 1*5 = 5. Write the factors and multiply.

(x+1)(x+5) = \(x^2 + 6x + 5\).

K = 6, Answer C.
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Let \(C\) and \(K\) be constants.If \(x^2+Kx+5\) factors into \((x+1)*(x+C)\), then what is the value of \(K\)?

\(x^2+Kx+5\) = \((x+1)*(x+C)\)

As we know that \(a^2 + 2ab + b^2\) = \((a+b) * (a+b)\)

After comparing the above equation with our equation we can see that

b = c = 5

We can also directly equate \(2ab = kx = 6x\)- However in order to make it more clear we will understand in detail per below steps.

\(x^2+Kx+5\) = \(x^2 + 5x + 1x + 5\)

\(x^2+Kx+5\) = \(x^2 + 6x + 5\)

\(kx = 6x\)

\(k = 6\)

Hence, the answer is C
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We know that x^2 + kx + 5 factors into (x + 1)(x + C)

We also know that k = sum of 1 + C, and that 5 = 1 * C --> C has to equal 5

And since k = 1 + C ---> k = 1 + 5

Just working with the formulas
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