Solving Quadratic Equations - GMAT Quantitative

The number of real solutions of the equation depends on whether discriminant is positive, zero, or negative; since , this becomes .

If we only know that is a perfect square, then we still need to know to find the number of real solutions. For example, let , a perfect square. Then the discriminant is , which can be positive, zero, or negative depending on .

But if we know , then the discriminant is

Therefore, has one real solution.

The sign of the discriminant of the quadratic expression answers this question; here, the discriminant is

,

or

If we assume Statement 1 alone, this expression becomes

Since we can assume is nonzero, . This makes the discriminant positive, proving that there are two real solutions.

If we assume Statement 2 alone, this expression becomes

The sign of can vary.

Case 1:

Then

giving the equation two imaginary solutions.

Case 2:

Then

giving the equation two real solutions.

Therefore, Statement 1, but not Statement 2, is enough to answer the question.

The two statements together are insufficient. If both are assumed, then can be 2, 4, or 6.

If the statement is true:

But if the statement is false:

The quadratic expression can be factored as , replacing the question marks with integers whose product is 8 and whose sum is . These integers are , so the equation becomes:

Set each linear binomial to 0 and solve:

Therefore, for the statement to be true, either or . Each of Statement 1 and Statement 2, taken alone, leaves other possible values of . Taken together, however, they are enough, since the only two positive even integers less than 6 are 2 and 4.

can be rewritten as

If Statement 1 holds, then the equation can be rewritten as . This equation has solution set , which comprises two real numbers.

If Statement 2 holds, the discriminant is positive, being the sum of a nonnegative number and a positive number; this makes the solution set one with two real numbers.

This is a quadratic formula problem. Use equation . For our problem, Plug these values into the equation, and simplify:

. Here, we simplified the radical by

The number of real solutions of the equation depends on whether discriminant is positive, zero, or negative; since , this becomes .

If we only know that is a perfect square, then we still need to know to find the number of real solutions. For example, let , a perfect square. Then the discriminant is , which can be positive, zero, or negative depending on .

But if we know , then the discriminant is

Therefore, has one real solution.

The sign of the discriminant of the quadratic expression answers this question; here, the discriminant is

,

or

If we assume Statement 1 alone, this expression becomes

Since we can assume is nonzero, . This makes the discriminant positive, proving that there are two real solutions.

If we assume Statement 2 alone, this expression becomes

The sign of can vary.

Case 1:

Then

giving the equation two imaginary solutions.

Case 2:

Then

giving the equation two real solutions.

Therefore, Statement 1, but not Statement 2, is enough to answer the question.

The two statements together are insufficient. If both are assumed, then can be 2, 4, or 6.

If the statement is true:

But if the statement is false:

The quadratic expression can be factored as , replacing the question marks with integers whose product is 8 and whose sum is . These integers are , so the equation becomes:

Set each linear binomial to 0 and solve:

Therefore, for the statement to be true, either or . Each of Statement 1 and Statement 2, taken alone, leaves other possible values of . Taken together, however, they are enough, since the only two positive even integers less than 6 are 2 and 4.

can be rewritten as

If Statement 1 holds, then the equation can be rewritten as . This equation has solution set , which comprises two real numbers.

If Statement 2 holds, the discriminant is positive, being the sum of a nonnegative number and a positive number; this makes the solution set one with two real numbers.

This is a quadratic formula problem. Use equation . For our problem, Plug these values into the equation, and simplify:

. Here, we simplified the radical by

The number of real solutions of the equation depends on whether discriminant is positive, zero, or negative; since , this becomes .

If we only know that is a perfect square, then we still need to know to find the number of real solutions. For example, let , a perfect square. Then the discriminant is , which can be positive, zero, or negative depending on .

But if we know , then the discriminant is

Therefore, has one real solution.

The sign of the discriminant of the quadratic expression answers this question; here, the discriminant is

,

or

If we assume Statement 1 alone, this expression becomes

Since we can assume is nonzero, . This makes the discriminant positive, proving that there are two real solutions.

If we assume Statement 2 alone, this expression becomes

The sign of can vary.

Case 1:

Then

giving the equation two imaginary solutions.

Case 2:

Then

giving the equation two real solutions.

Therefore, Statement 1, but not Statement 2, is enough to answer the question.

The two statements together are insufficient. If both are assumed, then can be 2, 4, or 6.

If the statement is true:

But if the statement is false:

The quadratic expression can be factored as , replacing the question marks with integers whose product is 8 and whose sum is . These integers are , so the equation becomes:

Set each linear binomial to 0 and solve:

Therefore, for the statement to be true, either or . Each of Statement 1 and Statement 2, taken alone, leaves other possible values of . Taken together, however, they are enough, since the only two positive even integers less than 6 are 2 and 4.

can be rewritten as

If Statement 1 holds, then the equation can be rewritten as . This equation has solution set , which comprises two real numbers.

If Statement 2 holds, the discriminant is positive, being the sum of a nonnegative number and a positive number; this makes the solution set one with two real numbers.

This is a quadratic formula problem. Use equation . For our problem, Plug these values into the equation, and simplify:

. Here, we simplified the radical by

The number of real solutions of the equation depends on whether discriminant is positive, zero, or negative; since , this becomes .

If we only know that is a perfect square, then we still need to know to find the number of real solutions. For example, let , a perfect square. Then the discriminant is , which can be positive, zero, or negative depending on .

But if we know , then the discriminant is

Therefore, has one real solution.