DSQ: Solving quadratic equations - GMAT Quantitative Reasoning

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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 $\sqrt{76}=\sqrt{4*19}=\sqrt{4}\sqrt{19}=2\sqrt{19}.$

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