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300 tough Algebra

by yellowho » Wed Jan 19, 2011 10:01 pm
What is the sum of all possible solutions for x of the equation x ( x − k ) = k + 1?

(A) 0
(B) 1
(C) k
(D) k + 1
(E) 2k - 1

This took a while to figure out. How are people solving this?
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by Anurag@Gurome » Wed Jan 19, 2011 10:19 pm
yellowho wrote:What is the sum of all possible solutions for x of the equation x ( x − k ) = k + 1?

(A) 0
(B) 1
(C) k
(D) k + 1
(E) 2k - 1
Any quadratic equations in x with two roots a and b can be expressed as, (x - a)(x - b) = 0. Which on expansion takes the form, x² - (a + b)x - ab = 0

Hence, the sum of the roots = -(coefficient of x in the expanded form)

Now, x(x − k) = k + 1
=> x² - kx - (k + 1) = 0

Hence, sum of the roots = -(-k) = k

The correct answer is C.
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by GMATGuruNY » Thu Jan 20, 2011 4:04 am
yellowho wrote:What is the sum of all possible solutions for x of the equation x ( x − k ) = k + 1?

(A) 0
(B) 1
(C) k
(D) k + 1
(E) 2k - 1

This took a while to figure out. How are people solving this?
The quickest approach might be to plug in a value for k.

Let k=2.
x(x-2) = 2+1
x^2 - 2x = 3
x^2 - 2x - 3 = 0.
(x-3)(x+1) = 0
x = 3, -1
Sum = 3+(-1) = 2. This is our target.

Now we plug k=2 into all the answer choices in order to see which yields our target of 2.

Only answer choice C works.

The correct answer is C.

Plugging in can be a very efficient way to determine the correct answer.
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by Everest » Thu Jan 20, 2011 1:38 pm
x ( x - k ) = k + 1, when wrote in quadratic form becomes, x^2 - Kx -(k+1)

find the roots of the above equation using formula -b+ sqrt(b^2 - 4ac/ 2a and -b- sqrt(b^2 - 4ac/ 2a .

so two solutions are k+1 and -1, add K+1 and -1 will result in K.


K is the answer
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by thebigkats » Sun Feb 20, 2011 3:53 pm
x ( x - k ) = k + 1
==> x^2 - k*x - (k+1) = 0
==> x^2 - (k+1)*x + 1*x - (k+1) = 0
==> (x+1) (x- (k+1)) = 0
==> x = -1 and x = k+1
==> SUM = -1+k+1
==> k


Of course there is a quicker solution by plugging values....
Anurag - you mentioned "Any quadratic equations in x with two roots a and b can be expressed as, (x - a)(x - b) = 0. Which on expansion takes the form, x² - (a + b)x - ab = 0 "
Did you mean to say "+ab"

Thanks,
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by stormier » Sun Feb 20, 2011 6:30 pm
thebigkats wrote:x ( x - k ) = k + 1
==> x^2 - k*x - (k+1) = 0
==> x^2 - (k+1)*x + 1*x - (k+1) = 0
==> (x+1) (x- (k+1)) = 0
==> x = -1 and x = k+1
==> SUM = -1+k+1
==> k


Of course there is a quicker solution by plugging values....
Anurag - you mentioned "Any quadratic equations in x with two roots a and b can be expressed as, (x - a)(x - b) = 0. Which on expansion takes the form, x² - (a + b)x - ab = 0 "
Did you mean to say "+ab"

Thanks,
Yes, that should be + ab. However that is irrelevant to the question which asks only about sum of roots and not the products.

the roots of a quadratic equation of the form

ax^2 + bx +c = 0

have the following properties:

(1) sum of roots = -b/a
(2) product of roots = c/a

roots = [-b +/- sqrt (b^2 - 4ac)]/2a

Add the two roots to derive (1)
Multiply the two roots - of the form (x+y)(x-y) - to derive (2)
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by thebigkats » Mon Feb 21, 2011 1:09 am
thanks stormier. Just wanted to clarify... agree that it has nothing to do with the question itself.
And thanks for the short cuts. I am sure they would be helpful

regards,


stormier wrote:
thebigkats wrote:x ( x - k ) = k + 1
==> x^2 - k*x - (k+1) = 0
==> x^2 - (k+1)*x + 1*x - (k+1) = 0
==> (x+1) (x- (k+1)) = 0
==> x = -1 and x = k+1
==> SUM = -1+k+1
==> k


Of course there is a quicker solution by plugging values....
Anurag - you mentioned "Any quadratic equations in x with two roots a and b can be expressed as, (x - a)(x - b) = 0. Which on expansion takes the form, x² - (a + b)x - ab = 0 "
Did you mean to say "+ab"

Thanks,
Yes, that should be + ab. However that is irrelevant to the question which asks only about sum of roots and not the products.

the roots of a quadratic equation of the form

ax^2 + bx +c = 0

have the following properties:

(1) sum of roots = -b/a
(2) product of roots = c/a

roots = [-b +/- sqrt (b^2 - 4ac)]/2a

Add the two roots to derive (1)
Multiply the two roots - of the form (x+y)(x-y) - to derive (2)
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by anirudhbhalotia » Thu Apr 28, 2011 7:01 pm
GMATGuruNY wrote:
yellowho wrote:What is the sum of all possible solutions for x of the equation x ( x − k ) = k + 1?

(A) 0
(B) 1
(C) k
(D) k + 1
(E) 2k - 1

This took a while to figure out. How are people solving this?
The quickest approach might be to plug in a value for k.

Let k=2.
x(x-2) = 2+1
x^2 - 2x = 3
x^2 - 2x - 3 = 0.
(x-3)(x+1) = 0
x = 3, -1
Sum = 3+(-1) = 2. This is our target.

Now we plug k=2 into all the answer choices in order to see which yields our target of 2.

Only answer choice C works.

The correct answer is C.

Plugging in can be a very efficient way to determine the correct answer.
What if we plug k as 3 or k as 1 or some other value?

You mentioned plugging in the sum in the answer choices...I didn't quiet understand this.
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by GMATGuruNY » Thu Apr 28, 2011 8:00 pm
anirudhbhalotia wrote:
GMATGuruNY wrote:
yellowho wrote:What is the sum of all possible solutions for x of the equation x ( x − k ) = k + 1?

(A) 0
(B) 1
(C) k
(D) k + 1
(E) 2k - 1

This took a while to figure out. How are people solving this?
The quickest approach might be to plug in a value for k.

Let k=2.
x(x-2) = 2+1
x^2 - 2x = 3
x^2 - 2x - 3 = 0.
(x-3)(x+1) = 0
x = 3, -1
Sum = 3+(-1) = 2. This is our target.

Now we plug k=2 into all the answer choices in order to see which yields our target of 2.

Only answer choice C works.

The correct answer is C.

Plugging in can be a very efficient way to determine the correct answer.
What if we plug k as 3 or k as 1 or some other value?

You mentioned plugging in the sum in the answer choices...I didn't quiet understand this.
Any value for k should work.
We shouldn't plug in k=1, however, because 1 is among the answer choices.
Avoid plugging in numbers that are in the answer choices.
The goal is to plug in a value that will make the math easy.

Let k=3:
x(x-3) = 3+1
x² - 3x = 4
x² - 3x - 4 = 0.
(x-4)(x+1) = 0
x = 4, -1
Sum = 4+(-1) = 3. This is our target.

Since we know that when k=3 the sum is 3, eliminate A and B.
Now we plug k=3 into C, D and E to see which yields our target of 3.

Answer choice C: k
k=3. This works.

Answer choice D: k+1
k+1 = 3+1 = 4. Doesn't work.

Answer choice E: 2k-1
2k-1 = 2*3 - 1 = 5. Doesn't work.

Once again, only answer choice C works.

The correct answer is C.
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by quantskillsgmat » Thu Apr 28, 2011 9:16 pm
answering of these typr of question shd nt be more than 10 seconds.
this is how
any qudratic equation has two zeroes(roots).
sum of roots is -b/a
products of roots is c/a.
use this get ur answer fast.
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by gmatisgay » Sat Aug 06, 2011 2:02 pm
hmmmm......however i think the answer should be D
this is how I did it
X(X-K)=K+1
X^2-XK=K+1
(X^2-1)=XK+K
(X+1)(X-1)=XK+K
(X+1)(X-1)=K(X+1)
cross out (X+1) then the solution leaves us (X-1)=k
therefore X=K+1

I am wrong??
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by Taniuca » Sat Aug 06, 2011 4:28 pm
I came to the same aswer as you gmatisgay "D". Could someone tell why this wouldn't be right? common factors to solve the equation should work for this type of problem.
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by goalevan » Sat Aug 06, 2011 4:29 pm
x ( x - k ) = k + 1
x^2 - kx - k - 1 = 0
x^2 - 1 - k(x + 1) = 0
(x + 1)(x - 1) - k(x + 1) = 0
(x + 1)[(x - 1) - k] = 0
(x + 1)(x - k - 1) = 0

So, either x + 1 = 0, or x = -1, or x - k - 1 = 0, or x = k + 1, (or both).

-1 + k + 1 = k
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by gnaneswari » Wed Feb 15, 2012 9:44 am
X(X-K)=K+1
X^2-XK=K+1
(X^2-1)=XK+K
(X+1)(X-1)=XK+K
(X+1)(X-1)=K(X+1)
cross out (X+1) then the solution leaves us (X-1)=k
therefore X=K+1

To emphasise, The question is 'what is the sum of all possible solutions of x'.

--> If you see the power of x in the given equation, it is 2,so quadratic equation. This means, there will be always two solutions/roots for x.
--> Your answer "X=k+1" is one of the solutions/roots of x.

If you subsititute 'x=-1' for the above equation, it satisfies it, which means -1 is another solution for x.

So sum of the of possible solutions, k+1 + (-1) = K.
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by ronnie1985 » Wed Apr 11, 2012 9:09 am
Reduce it to a quadratic equation. Sum of roots is -b/a, in this case, k
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