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graceo
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About k^2+k-6 22 Jul 2020, 16:33
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We know that k^2+k-6 = (k-3)(k+2)

I see that we are diving "-6" into "+/-3" times "+/-2". However, can anyone tell me why is it "-3 and 2", instead of "3 and -2"?

Thanks!

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About k^2+k-6 22 Jul 2020, 19:37
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graceo
We know that k^2+k-6 = (k-3)(k+2)

I see that we are diving "-6" into "+/-3" times "+/-2". However, can anyone tell me why is it "-3 and 2", instead of "3 and -2"?

Thanks!
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Your factorisation is wrong.

k^2+k-6 = k^2+(3k-2k)-6 = (k+3)(k-2)

Hence, roots of the equation are -3 and +2
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CrackverbalGMAT
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About k^2+k-6 22 Jul 2020, 19:38
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graceo
We know that k^2+k-6 = (k-3)(k+2)

I see that we are diving "-6" into "+/-3" times "+/-2". However, can anyone tell me why is it "-3 and 2", instead of "3 and -2"?

Thanks!
Show more

When finding the roots of the quadratic equation a\(x^2\) + bx + x = 0, using the method of splitting the middle term into 2 parts, then

(i) The sum of the middle terms should be = b

(ii) The product of the middle terms should be ac

In \(k^2\) + k - 6, the split should happen such that the sum of the terms = 1 and the product of the terms = 1*-6 = -6

Therefore the middle terms should be +3k and -2k


We get \(k^2\) + 3k - 2k - 6 = 0

k(k + 3) - 2(k + 3) = 0

(k + 3)(k - 2) = 0

The roots are therefore -3 and 2

Hope This Helps

Arun Kumar
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About k^2+k-6 22 Jul 2020, 21:40
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graceo
We know that k^2+k-6 = (k-3)(k+2)

I see that we are diving "-6" into "+/-3" times "+/-2". However, can anyone tell me why is it "-3 and 2", instead of "3 and -2"?

Thanks!
Show more

You can take out smaller power only as common.
example
\(X^2 +x^3\)
can be written as \(x^2(x+1)\)
similarly -6 is smaller hence common out
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graceo
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About k^2+k-6 27 Jul 2020, 20:59
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Thank you!! This makes so much sense:)

CrackVerbalGMAT
graceo
We know that k^2+k-6 = (k-3)(k+2)

I see that we are diving "-6" into "+/-3" times "+/-2". However, can anyone tell me why is it "-3 and 2", instead of "3 and -2"?

Thanks!

When finding the roots of the quadratic equation a\(x^2\) + bx + x = 0, using the method of splitting the middle term into 2 parts, then

(i) The sum of the middle terms should be = b

(ii) The product of the middle terms should be ac

In \(k^2\) + k - 6, the split should happen such that the sum of the terms = 1 and the product of the terms = 1*-6 = -6

Therefore the middle terms should be +3k and -2k


We get \(k^2\) + 3k - 2k - 6 = 0

k(k + 3) - 2(k + 3) = 0

(k + 3)(k - 2) = 0

The roots are therefore -3 and 2

Hope This Helps

Arun Kumar
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graceo
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About k^2+k-6 27 Jul 2020, 21:00
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Yup I realized lol...thank you!

pranav952


Your factorisation is wrong.

k^2+k-6 = k^2+(3k-2k)-6 = (k+3)(k-2)

Hence, roots of the equation are -3 and +2
Show more

Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!