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Updated on: 06 Jul 2019, 04:25
Originally posted by stonecold on 22 Apr 2017, 12:02.
Last edited by Sajjad1994 on 06 Jul 2019, 04:25, edited 1 time in total.
Last edited by Sajjad1994 on 06 Jul 2019, 04:25, edited 1 time in total.
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Difficulty:
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Question Stats:
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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
A)0
B)5
C)6
D)8
E)Cannot be determined
Source: Nova GMAT
Difficulty Level: 550
Difficulty Level: 550
Matching constants value on the given equation with its factor ... we get C=5 and K=c+1=6
Option C
Posted from my mobile device
Option C
Posted from my mobile device
Posts: 17
we have been given that X^2+kx+5 in to factor equtations (x+1) (x+c)
now solve this (x+1) (x+c) = x^2+x+cx+c
x^2 +x(1+c) + c
Now we know that equation is x^2 +kx+5 form so compare this equation with x^2 +X(1+c) + C
we will comclude that c=5, kx will be = x(1+c) i.e. 6x since c=5
so the value of k=6
now solve this (x+1) (x+c) = x^2+x+cx+c
x^2 +x(1+c) + c
Now we know that equation is x^2 +kx+5 form so compare this equation with x^2 +X(1+c) + C
we will comclude that c=5, kx will be = x(1+c) i.e. 6x since c=5
so the value of k=6
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
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
stonecold
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
\(x^2+Kx+5\) = \((x+1)*(x+C)\)
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
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
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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