DSQ: Solving quadratic equations - GMAT Quantitative

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Question

What are the solutions of $5x^{2}-4x-3$ in the most simplified form?

Answer

This is a quadratic formula problem. Use equation $\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ . For our problem, Plug these values into the equation, and simplify:

$\frac{2 \pm \sqrt{(-4)^{2} - 4(5)(-3)}}{2(5)}$ $=\frac{2 \pm \sqrt{16 +60}}{10}=\frac{2 \pm \sqrt{76}}{10}=\frac{1 \pm \sqrt{19}}{5}$ . Here, we simplified the radical by $\sqrt{76}=\sqrt{4*19}=\sqrt{4}\sqrt{19}=2\sqrt{19}.$

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Question

Consider the equation $x^{2} +Bx = C$

How many real solutions does this equation have?

Statement 1: There exists two different real numbers such that and

Statement 2: is a positive integer.

Answer

$x^{2} +Bx = C$ can be rewritten as $x^{2} +Bx - C = 0$

If Statement 1 holds, then the equation can be rewritten as . This equation has solution set $\left { -m,-n\right }$ , which comprises two real numbers.

If Statement 2 holds, the discriminant $B^{2} -4 \cdot1\cdot(-C)= B^{2} +4C$ is positive, being the sum of a nonnegative number and a positive number; this makes the solution set one with two real numbers.

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Question

Let be two positive integers. How many real solutions does the equation $x^{2} + Bx + C = 0$ have?

Statement 1: is a perfect square of an integer.

Statement 2: $C = \left ( \frac{B}{2} \right ) ^{2}$

Answer

The number of real solutions of the equation $Ax^{2} + Bx + C = 0$ depends on whether discriminant $B^{2} - 4A C$ is positive, zero, or negative; since , this becomes $B^{2} - 4C$ .

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 $B^{2} - 4$ , which can be positive, zero, or negative depending on .

But if we know $C = \left ( \frac{B}{2} \right ) ^{2}$ , then the discriminant is

$B^{2} - 4C = B^{2} - 4 \cdot \left ( \frac{B}{2} \right ) ^{2} = B^{2} - 4 \cdot \frac{B^{2}}{4} = B^{2} - B^{2} = 0$

Therefore, $x^{2} + Bx + C = 0$ has one real solution.

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Question

Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?

$Ax^{2} -3Ax + B = 0$

Statement 1:

Statement 2:

Answer

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

$(-3A)^{2} - 4AB$ ,

or

$9A^{2} - 4AB$

If we assume Statement 1 alone, this expression becomes

$9A^{2} - 4AB = 9A^{2} - 4A (2A) = 9A^{2} - 8A^{2} = A^{2}$

Since we can assume is nonzero, $A^{2} > 0$ . This makes the discriminant positive, proving that there are two real solutions.

If we assume Statement 2 alone, this expression becomes

$9A^{2} - 4AB = 9A^{2} - 4A \left ( A + 2\right ) = 9A^{2} - 4A^{2}-8A = 5A^{2}-8A$

The sign of $5A^{2}-8A$ can vary.

Case 1:

Then $5A^{2}-8A = 5 \cdot 1^{2}-8\cdot 1 = 5 - 8 = -3 < 0$

giving the equation two imaginary solutions.

Case 2:

Then $5A^{2}-8A = 5 \cdot 2^{2}-8\cdot 2 = 20 - 16 = 4 > 0$

giving the equation two real solutions.

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

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Question

is a positive integer.

True or false?

$x^{2} - 10x +24 = 0$

Statement 1: is an even integer

Statement 2:

Answer

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

If the statement is true:

$x^{2} - 10x +24 = 0$

$4^{2} - 10 \cdot 4 +24 = 0$

But if the statement is false:

$x^{2} - 10x +24 = 0$

$2^{2} - 10 \cdot 2 +24 = 0$

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Question

is a positive integer.

True or false:

$x^{2} - 6x + 8 = 0$

Statement 1: is an even integer

Statement 2:

Answer

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:

$x^{2} - 6x + 8 = 0$

Set each linear binomial to 0 and solve:

$x - 2 = 0 \Rightarrow x = 2$

$x - 4 = 0 \Rightarrow x = 4$

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.

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Question

Consider the equation $x^{2} +Bx = C$

How many real solutions does this equation have?

Statement 1: There exists two different real numbers such that and

Statement 2: is a positive integer.

Answer

$x^{2} +Bx = C$ can be rewritten as $x^{2} +Bx - C = 0$

If Statement 1 holds, then the equation can be rewritten as . This equation has solution set $\left { -m,-n\right }$ , which comprises two real numbers.

If Statement 2 holds, the discriminant $B^{2} -4 \cdot1\cdot(-C)= B^{2} +4C$ is positive, being the sum of a nonnegative number and a positive number; this makes the solution set one with two real numbers.

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Question

Let be two positive integers. How many real solutions does the equation $x^{2} + Bx + C = 0$ have?

Statement 1: is a perfect square of an integer.

Statement 2: $C = \left ( \frac{B}{2} \right ) ^{2}$

Answer

The number of real solutions of the equation $Ax^{2} + Bx + C = 0$ depends on whether discriminant $B^{2} - 4A C$ is positive, zero, or negative; since , this becomes $B^{2} - 4C$ .

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 $B^{2} - 4$ , which can be positive, zero, or negative depending on .

But if we know $C = \left ( \frac{B}{2} \right ) ^{2}$ , then the discriminant is

$B^{2} - 4C = B^{2} - 4 \cdot \left ( \frac{B}{2} \right ) ^{2} = B^{2} - 4 \cdot \frac{B^{2}}{4} = B^{2} - B^{2} = 0$

Therefore, $x^{2} + Bx + C = 0$ has one real solution.

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Question

Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?

$Ax^{2} -3Ax + B = 0$

Statement 1:

Statement 2:

Answer

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

$(-3A)^{2} - 4AB$ ,

or

$9A^{2} - 4AB$

If we assume Statement 1 alone, this expression becomes

$9A^{2} - 4AB = 9A^{2} - 4A (2A) = 9A^{2} - 8A^{2} = A^{2}$

Since we can assume is nonzero, $A^{2} > 0$ . This makes the discriminant positive, proving that there are two real solutions.

If we assume Statement 2 alone, this expression becomes

$9A^{2} - 4AB = 9A^{2} - 4A \left ( A + 2\right ) = 9A^{2} - 4A^{2}-8A = 5A^{2}-8A$

The sign of $5A^{2}-8A$ can vary.

Case 1:

Then $5A^{2}-8A = 5 \cdot 1^{2}-8\cdot 1 = 5 - 8 = -3 < 0$

giving the equation two imaginary solutions.

Case 2:

Then $5A^{2}-8A = 5 \cdot 2^{2}-8\cdot 2 = 20 - 16 = 4 > 0$

giving the equation two real solutions.

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

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Question

is a positive integer.

True or false?

$x^{2} - 10x +24 = 0$

Statement 1: is an even integer

Statement 2:

Answer

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

If the statement is true:

$x^{2} - 10x +24 = 0$

$4^{2} - 10 \cdot 4 +24 = 0$

But if the statement is false:

$x^{2} - 10x +24 = 0$

$2^{2} - 10 \cdot 2 +24 = 0$

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Question

is a positive integer.

True or false:

$x^{2} - 6x + 8 = 0$

Statement 1: is an even integer

Statement 2:

Answer

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:

$x^{2} - 6x + 8 = 0$

Set each linear binomial to 0 and solve:

$x - 2 = 0 \Rightarrow x = 2$

$x - 4 = 0 \Rightarrow x = 4$

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.

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Question

What are the solutions of $5x^{2}-4x-3$ in the most simplified form?

Answer

This is a quadratic formula problem. Use equation $\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ . For our problem, Plug these values into the equation, and simplify:

$\frac{2 \pm \sqrt{(-4)^{2} - 4(5)(-3)}}{2(5)}$ $=\frac{2 \pm \sqrt{16 +60}}{10}=\frac{2 \pm \sqrt{76}}{10}=\frac{1 \pm \sqrt{19}}{5}$ . Here, we simplified the radical by $\sqrt{76}=\sqrt{4*19}=\sqrt{4}\sqrt{19}=2\sqrt{19}.$

Compare your answer with the correct one above

Question

Consider the equation $x^{2} +Bx = C$

How many real solutions does this equation have?

Statement 1: There exists two different real numbers such that and

Statement 2: is a positive integer.

Answer

$x^{2} +Bx = C$ can be rewritten as $x^{2} +Bx - C = 0$

If Statement 1 holds, then the equation can be rewritten as . This equation has solution set $\left { -m,-n\right }$ , which comprises two real numbers.

If Statement 2 holds, the discriminant $B^{2} -4 \cdot1\cdot(-C)= B^{2} +4C$ is positive, being the sum of a nonnegative number and a positive number; this makes the solution set one with two real numbers.

Compare your answer with the correct one above

Question

Let be two positive integers. How many real solutions does the equation $x^{2} + Bx + C = 0$ have?

Statement 1: is a perfect square of an integer.

Statement 2: $C = \left ( \frac{B}{2} \right ) ^{2}$

Answer

The number of real solutions of the equation $Ax^{2} + Bx + C = 0$ depends on whether discriminant $B^{2} - 4A C$ is positive, zero, or negative; since , this becomes $B^{2} - 4C$ .

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 $B^{2} - 4$ , which can be positive, zero, or negative depending on .

But if we know $C = \left ( \frac{B}{2} \right ) ^{2}$ , then the discriminant is

$B^{2} - 4C = B^{2} - 4 \cdot \left ( \frac{B}{2} \right ) ^{2} = B^{2} - 4 \cdot \frac{B^{2}}{4} = B^{2} - B^{2} = 0$

Therefore, $x^{2} + Bx + C = 0$ has one real solution.

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Question

Does the solution set of the following quadratic equation comprise two real solutions, one real solution, or one imaginary solution?

$Ax^{2} -3Ax + B = 0$

Statement 1:

Statement 2:

Answer

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

$(-3A)^{2} - 4AB$ ,

or

$9A^{2} - 4AB$

If we assume Statement 1 alone, this expression becomes

$9A^{2} - 4AB = 9A^{2} - 4A (2A) = 9A^{2} - 8A^{2} = A^{2}$

Since we can assume is nonzero, $A^{2} > 0$ . This makes the discriminant positive, proving that there are two real solutions.

If we assume Statement 2 alone, this expression becomes

$9A^{2} - 4AB = 9A^{2} - 4A \left ( A + 2\right ) = 9A^{2} - 4A^{2}-8A = 5A^{2}-8A$

The sign of $5A^{2}-8A$ can vary.

Case 1:

Then $5A^{2}-8A = 5 \cdot 1^{2}-8\cdot 1 = 5 - 8 = -3 < 0$

giving the equation two imaginary solutions.

Case 2:

Then $5A^{2}-8A = 5 \cdot 2^{2}-8\cdot 2 = 20 - 16 = 4 > 0$

giving the equation two real solutions.

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

Compare your answer with the correct one above

Question

is a positive integer.

True or false?

$x^{2} - 10x +24 = 0$

Statement 1: is an even integer

Statement 2:

Answer

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

If the statement is true:

$x^{2} - 10x +24 = 0$

$4^{2} - 10 \cdot 4 +24 = 0$

But if the statement is false:

$x^{2} - 10x +24 = 0$

$2^{2} - 10 \cdot 2 +24 = 0$

Compare your answer with the correct one above

Question

is a positive integer.

True or false:

$x^{2} - 6x + 8 = 0$

Statement 1: is an even integer

Statement 2:

Answer

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:

$x^{2} - 6x + 8 = 0$

Set each linear binomial to 0 and solve:

$x - 2 = 0 \Rightarrow x = 2$

$x - 4 = 0 \Rightarrow x = 4$

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.

Compare your answer with the correct one above

Question

What are the solutions of $5x^{2}-4x-3$ in the most simplified form?

Answer

This is a quadratic formula problem. Use equation $\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ . For our problem, Plug these values into the equation, and simplify:

$\frac{2 \pm \sqrt{(-4)^{2} - 4(5)(-3)}}{2(5)}$ $=\frac{2 \pm \sqrt{16 +60}}{10}=\frac{2 \pm \sqrt{76}}{10}=\frac{1 \pm \sqrt{19}}{5}$ . Here, we simplified the radical by $\sqrt{76}=\sqrt{4*19}=\sqrt{4}\sqrt{19}=2\sqrt{19}.$

Compare your answer with the correct one above

Question

Consider the equation $x^{2} +Bx = C$

How many real solutions does this equation have?

Statement 1: There exists two different real numbers such that and

Statement 2: is a positive integer.

Answer

$x^{2} +Bx = C$ can be rewritten as $x^{2} +Bx - C = 0$

If Statement 1 holds, then the equation can be rewritten as . This equation has solution set $\left { -m,-n\right }$ , which comprises two real numbers.

If Statement 2 holds, the discriminant $B^{2} -4 \cdot1\cdot(-C)= B^{2} +4C$ is positive, being the sum of a nonnegative number and a positive number; this makes the solution set one with two real numbers.

Compare your answer with the correct one above

Question

Let be two positive integers. How many real solutions does the equation $x^{2} + Bx + C = 0$ have?

Statement 1: is a perfect square of an integer.

Statement 2: $C = \left ( \frac{B}{2} \right ) ^{2}$

Answer

The number of real solutions of the equation $Ax^{2} + Bx + C = 0$ depends on whether discriminant $B^{2} - 4A C$ is positive, zero, or negative; since , this becomes $B^{2} - 4C$ .

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 $B^{2} - 4$ , which can be positive, zero, or negative depending on .

But if we know $C = \left ( \frac{B}{2} \right ) ^{2}$ , then the discriminant is

$B^{2} - 4C = B^{2} - 4 \cdot \left ( \frac{B}{2} \right ) ^{2} = B^{2} - 4 \cdot \frac{B^{2}}{4} = B^{2} - B^{2} = 0$

Therefore, $x^{2} + Bx + C = 0$ has one real solution.

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