For all integers n, n* = n(n – 1). What is the value of x* w

quick question, when (x-1)(x-2) = (x - 2) why cant we cross out (x-2) and arrive at x-1=1 and x=2 ?

As a general rule, never cross multiply variables on two sides of an equation unless we can be sure that it cannot be equal to 0.

In this case it turned out that the answer would have come the same. However, there is no way of knowing that in advance.
Take the following case :
(x-2)(x-2) = (x-2) ----> if we cross out (x-2) on both sides, we will get x = 3. However, x = 2 also satisfies this equation.
(x-3)(x-2) = (x-2) ----> again if we cross out (x-2) on both sides we will get x = 4. However, x = 2 also satisfies this equation.

One more thing to note is that an algebraic expression, the highest power of x will determine the number of roots of x. Therefore a quadratic equation in x will always have 2 values that satisfy it. It can be that the two values are the same as they are in our case. However, they will still be treated as two values. Thus we will say that the roots of the equation are 2 and 2.

One little thing: in quadratic equation highest power of x will determine maximum # of roots. Quadratic equation can have 0, 1, or 2 roots.

When discriminant more than 0 you'll have 2 roots;
When discriminant is zero you'll have 1 root (and not two treated as one);
When discriminant is less than 0 you won't have any real roots.

x^2-4x+3=0 has 2 roots;
x^2-4x+4=0 has 1 root;
x^2-4x+5=0 has 0 real roots.

In cases when the discriminant is more than 0 or less than 0, you will always have two distinct roots. In one case real and in the other a pair of complex conjugates.

What you have stated is perfectly valid provided we specify that we are only concerned with the real and distinct roots of an equation.

When the discriminant is 0, you are right in saying that it will have only one distinct root. However, the term root by itself does not imply distinct or real. Thus when the discriminant is 0 it actually has a double root (which is to account for its multiplicity).

So, unless we are asked to find the number of distinct real solutions of a quadratic equation am I not right in saying that it will always have two roots?

When the discriminant is negative quadratic equation has no real roots. As I stated above.

GMAT is dealing ONLY with real numbers. No need to complicate this.

When the quadratic equation has one root it's rarely called "double root". More common to say that it has 1 root. So usually when we say root of equation we think about the distinct root, even if we don't specify it.

So we can say:
Discriminant positive - 2 roots;
Discriminant 0 - 1 root, even not specifying that it's distinct;
Discriminant negative - no real roots, for GMAT no root at all.

Though you are right in saying that quadratic equation always has two roots if:
A. We consider complex roots, which is not the case for GMAT;
B. We consider the concept of "double root", though I've never seen GMAT even mentioning the double root.

From my experience in GMAT, we can say that it's considering quadratic equation as parabola and the roots as its intersection with X-axis. One tangent point - one root, two intersections - 2 roots, no intersection - no root.

Again as I said no need to complicate.

For all integers n, n* = n(n – 1). what is the value of x* when x is an integer?

(1) x* = x

(2) (x – 1)* = x – 2

My questions is whether you are algebraically allowed to divide by X if you have the equation: x(x-1) = x. Dividing by x yields x-1 = 1 whereas not dividing yields x^2 - 2x.

Changes the answer...

Thanks!

For all integers n, n* = n(n – 1). what is the value of x* when x is an integer?

(1) x* = x --> \(x(x-1)=x\) --> \(x(x-1)-x=0\) --> \(x(x-2)=0\) --> \(x=0\) or \(x=2\). Thus \(x*=0(0-1)=0\) or \(x*=2(2-1)=2\). Not sufficient.

If you divide (reduce) \(x(x-1)=x\) by x you assume, with no ground for it, that x does not equal to zero thus exclude a possible solution (notice that both x=0 AND x=2 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.


(2) (x – 1)* = x – 2 --> \((x-1)(x-2)=x-2\) --> \((x-2)^2=0\) --> \(x=2\). Sufficient.

Answer: B.

Hope it's clear.

HI Bunuel,

How are you getting (x-2)^2 = 0 in second statemnt?

Thanks.

\((x-1)(x-2)=x-2\);

\((x-1)(x-2)-(x-2)=0\);

\((x-2)(x-1-1)=0\);

\((x-2)(x-2)=0\).

Hope it's clear.

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