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.

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

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,

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)

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)

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.

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.

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.

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??

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.

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

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.

Reduce it to a quadratic equation. Sum of roots is -b/a, in this case, k