Solving Quadratic Equations - GMAT Quantitative
Card 0 of 48
What are the solutions of 
in the most simplified form?
What are the solutions of 
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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 
This is a quadratic formula problem. Use equation 
Let 
be two positive integers. How many real solutions does the equation 
have?
Statement 1: 
is a perfect square of an integer.
Statement 2: 
Let
Statement 1:
Statement 2:
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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 number of real solutions of the equation
If we only know that
But if we know
Therefore,
Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?
Statement 1: 
Statement 2: 
Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?
Statement 1:
Statement 2:
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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 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
If we assume Statement 2 alone, this expression becomes
The sign of
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.
is a positive integer.
True or false?
Statement 1: 
is an even integer
Statement 2: 
True or false?
Statement 1:
Statement 2:
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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 two statements together are insufficient. If both are assumed, then
If
But if
is a positive integer.
True or false:
Statement 1: is an even integer
Statement 2: 
True or false:
Statement 1:
Statement 2:
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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.
The quadratic expression can be factored as
Set each linear binomial to 0 and solve:
Therefore, for the statement to be true, either
What are the solutions of in the most simplified form?
What are the solutions of
Tap to see back →
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
This is a quadratic formula problem. Use equation
Let be two positive integers. How many real solutions does the equation have?
Statement 1: is a perfect square of an integer.
Statement 2:
Let
Statement 1:
Statement 2:
Tap to see back →
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 number of real solutions of the equation
If we only know that
But if we know
Therefore,
Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?
Statement 1:
Statement 2:
Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?
Statement 1:
Statement 2:
Tap to see back →
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 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
If we assume Statement 2 alone, this expression becomes
The sign of
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.
is a positive integer.
True or false?
Statement 1: is an even integer
Statement 2:
True or false?
Statement 1:
Statement 2:
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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 two statements together are insufficient. If both are assumed, then
If
But if
is a positive integer.
True or false:
Statement 1: is an even integer
Statement 2:
True or false:
Statement 1:
Statement 2:
Tap to see back →
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.
The quadratic expression can be factored as
Set each linear binomial to 0 and solve:
Therefore, for the statement to be true, either
What are the solutions of in the most simplified form?
What are the solutions of
Tap to see back →
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
This is a quadratic formula problem. Use equation
Let be two positive integers. How many real solutions does the equation have?
Statement 1: is a perfect square of an integer.
Statement 2:
Let
Statement 1:
Statement 2:
Tap to see back →
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 number of real solutions of the equation
If we only know that
But if we know
Therefore,
Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?
Statement 1:
Statement 2:
Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?
Statement 1:
Statement 2:
Tap to see back →
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 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
If we assume Statement 2 alone, this expression becomes
The sign of
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.
is a positive integer.
True or false?
Statement 1: is an even integer
Statement 2:
True or false?
Statement 1:
Statement 2:
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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 two statements together are insufficient. If both are assumed, then
If
But if
is a positive integer.
True or false:
Statement 1: is an even integer
Statement 2:
True or false:
Statement 1:
Statement 2:
Tap to see back →
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.
The quadratic expression can be factored as
Set each linear binomial to 0 and solve:
Therefore, for the statement to be true, either
What are the solutions of in the most simplified form?
What are the solutions of
Tap to see back →
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
This is a quadratic formula problem. Use equation
Let be two positive integers. How many real solutions does the equation have?
Statement 1: is a perfect square of an integer.
Statement 2:
Let
Statement 1:
Statement 2:
Tap to see back →
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 number of real solutions of the equation
If we only know that
But if we know
Therefore,
Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?
Statement 1:
Statement 2:
Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?
Statement 1:
Statement 2:
Tap to see back →
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 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
If we assume Statement 2 alone, this expression becomes
The sign of
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.
is a positive integer.
True or false?
Statement 1: is an even integer
Statement 2:
True or false?
Statement 1:
Statement 2:
Tap to see back →
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 two statements together are insufficient. If both are assumed, then
If
But if
is a positive integer.
True or false:
Statement 1: is an even integer
Statement 2:
True or false:
Statement 1:
Statement 2:
Tap to see back →
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.
The quadratic expression can be factored as
Set each linear binomial to 0 and solve:
Therefore, for the statement to be true, either