Equations - GMAT Quantitative

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Question

Give the solution set of the equation:

$\left | x ^{2} - 2x \right | = 3x -6$

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Answer

Write this as a compound equation and solve each separately.

$\left | x ^{2} - 2x \right | = 3x -6$

$x ^{2} - 2x = 3x -6 \textrm{ or }x ^{2} - 2x = -\left ( 3x -6 \right )$

$x ^{2} - 2x = 3x -6$

$x ^{2} - 2x - 3x + 6 = 3x -6 - 3x + 6$

$x ^{2} - 5x + 6 =0$

$x - 3 = 0 \Rightarrow x = 3$

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

$x ^{2} - 2x = -\left ( 3x -6 \right )$

$x ^{2} - 2x = -3x +6$

$x ^{2} - 2x+ 3x - 6= -3x +6 + 3x - 6$

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

$x + 3 = 0 \Rightarrow x = -3$

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

This gives us three possible solutions - $\left \{ -3, 2, 3\right \}$ . We check all three.

$\left | x ^{2} - 2x \right | = 3x -6$

$\left | \left ( -3\right ) ^{2} - 2 (-3) \right | = 3 (-3) -6$

$\left | \left9 - (-6) \right | = -9-6$

$\left |15| = -15$

This is a false statement so we can eliminate as a false "solution".

$\left | x ^{2} - 2x \right | = 3x -6$

$\left |2^{2} - 2 \cdot 2 \right | = 2 \cdot 3 -6$

$\left | \left 4-4 \right | = 6-6$

2 is a solution.

$\left | x ^{2} - 2x \right | = 3x -6$

$\left | 3 ^{2} - 2 \cdot 3 \right | = 3 \cdot 3 -6$

$\left | \left 9 - 6\right | = 9-6$

$\left |3 | = 3$

3 is a solution.

The solution set is $x \in \left \{ 2, 3\right \}$ .

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Question

Solve for , giving all solutions, real and imaginary:

$x^{3} - 4x^{2}+ 3x - 12 = 0$

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Answer

Factor the expression:

$x^{3} - 4x^{2}+ 3x - 12 = 0$

$\left ( x^{3} - 4x^{2} \right ) +\left ( 3x - 12 \right )= 0$

$x^{2} \left ( x - 4 \right ) + 3 \left ( x - 4 \right )= 0$

$\left (x^{2}+ 3 \right ) \left ( x - 4 \right ) = 0$

Rewrite:

or

$x^{2} +3 =0$

$x^{2} = -3$

$x = \pm \sqrt{ -3} = \pm i \sqrt{3}$

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Question

Which of the following is the solution set of the equation:

$x-3 = \sqrt{2x+9}$

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Answer

Square both sides of the equation, then solve the resulting quadratic equation by factoring:

$x-3 = \sqrt{2x+9}$

$\left (x-3 \right ) ^{2}=\left ( \sqrt{2x+9 \right )}^{2}$

$x^{2}-6x+9=2x+9$

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

Either

or

Checking both solutions, however:

$0-3 = \sqrt{2\cdot 0+9}$

This eliminates 0 as a solution.

$8-3 = \sqrt{2\cdot 8+9}$

8 turns out to be the only solution.

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Question

Define $f (x) = x \cdot |x |$ . Which of the following would be a valid alternative way of expressing the definition of ?

Tap to reveal answer

Answer

By definition:

If $x \geq 0$ , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot x= x^{2}$

If , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot \left (- x \right )= - x^{2}$

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Question

Which of the following is the solution set of the equation:

$x-3 = \sqrt{2x+9}$

Tap to reveal answer

Answer

Square both sides of the equation, then solve the resulting quadratic equation by factoring:

$x-3 = \sqrt{2x+9}$

$\left (x-3 \right ) ^{2}=\left ( \sqrt{2x+9 \right )}^{2}$

$x^{2}-6x+9=2x+9$

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

Either

or

Checking both solutions, however:

$0-3 = \sqrt{2\cdot 0+9}$

This eliminates 0 as a solution.

$8-3 = \sqrt{2\cdot 8+9}$

8 turns out to be the only solution.

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Question

Define $f (x) = x \cdot |x |$ . Which of the following would be a valid alternative way of expressing the definition of ?

Tap to reveal answer

Answer

By definition:

If $x \geq 0$ , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot x= x^{2}$

If , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot \left (- x \right )= - x^{2}$

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Question

Give the solution set of the equation:

$x^{2} =|x| + 6$

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Answer

Case 1: is nonnegative - that is, .

The equation becomes

$x^{2} =x+ 6$

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

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

Either , in which case ,

or , in which case , which is impossible, as we are assuming that is nonnegative.

case 2: is negative - that is,

The equation becomes

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

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

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

Either , in which case , which is impossible since we are assuming that is negative,

or , in which case .

and can both be confirmed as solutions.

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Question

Give the solution set of the equation:

$\left | x ^{2} - 2x \right | = 3x -6$

Tap to reveal answer

Answer

Write this as a compound equation and solve each separately.

$\left | x ^{2} - 2x \right | = 3x -6$

$x ^{2} - 2x = 3x -6 \textrm{ or }x ^{2} - 2x = -\left ( 3x -6 \right )$

$x ^{2} - 2x = 3x -6$

$x ^{2} - 2x - 3x + 6 = 3x -6 - 3x + 6$

$x ^{2} - 5x + 6 =0$

$x - 3 = 0 \Rightarrow x = 3$

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

$x ^{2} - 2x = -\left ( 3x -6 \right )$

$x ^{2} - 2x = -3x +6$

$x ^{2} - 2x+ 3x - 6= -3x +6 + 3x - 6$

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

$x + 3 = 0 \Rightarrow x = -3$

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

This gives us three possible solutions - $\left \{ -3, 2, 3\right \}$ . We check all three.

$\left | x ^{2} - 2x \right | = 3x -6$

$\left | \left ( -3\right ) ^{2} - 2 (-3) \right | = 3 (-3) -6$

$\left | \left9 - (-6) \right | = -9-6$

$\left |15| = -15$

This is a false statement so we can eliminate as a false "solution".

$\left | x ^{2} - 2x \right | = 3x -6$

$\left |2^{2} - 2 \cdot 2 \right | = 2 \cdot 3 -6$

$\left | \left 4-4 \right | = 6-6$

2 is a solution.

$\left | x ^{2} - 2x \right | = 3x -6$

$\left | 3 ^{2} - 2 \cdot 3 \right | = 3 \cdot 3 -6$

$\left | \left 9 - 6\right | = 9-6$

$\left |3 | = 3$

3 is a solution.

The solution set is $x \in \left \{ 2, 3\right \}$ .

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Question

Define $f (x) = x \cdot |x |$ . Which of the following would be a valid alternative way of expressing the definition of ?

Tap to reveal answer

Answer

By definition:

If $x \geq 0$ , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot x= x^{2}$

If , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot \left (- x \right )= - x^{2}$

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Question

Which of the following is the solution set of the equation:

$x-3 = \sqrt{2x+9}$

Tap to reveal answer

Answer

Square both sides of the equation, then solve the resulting quadratic equation by factoring:

$x-3 = \sqrt{2x+9}$

$\left (x-3 \right ) ^{2}=\left ( \sqrt{2x+9 \right )}^{2}$

$x^{2}-6x+9=2x+9$

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

Either

or

Checking both solutions, however:

$0-3 = \sqrt{2\cdot 0+9}$

This eliminates 0 as a solution.

$8-3 = \sqrt{2\cdot 8+9}$

8 turns out to be the only solution.

← Didn't Know|Knew It →

Question

Which of the following is the solution set of the equation:

$x-3 = \sqrt{2x+9}$

Tap to reveal answer

Answer

Square both sides of the equation, then solve the resulting quadratic equation by factoring:

$x-3 = \sqrt{2x+9}$

$\left (x-3 \right ) ^{2}=\left ( \sqrt{2x+9 \right )}^{2}$

$x^{2}-6x+9=2x+9$

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

Either

or

Checking both solutions, however:

$0-3 = \sqrt{2\cdot 0+9}$

This eliminates 0 as a solution.

$8-3 = \sqrt{2\cdot 8+9}$

8 turns out to be the only solution.

← Didn't Know|Knew It →

Question

Define $f (x) = x \cdot |x |$ . Which of the following would be a valid alternative way of expressing the definition of ?

Tap to reveal answer

Answer

By definition:

If $x \geq 0$ , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot x= x^{2}$

If , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot \left (- x \right )= - x^{2}$

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Question

Give the solution set of the equation:

$x^{2} =|x| + 6$

Tap to reveal answer

Answer

Case 1: is nonnegative - that is, .

The equation becomes

$x^{2} =x+ 6$

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

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

Either , in which case ,

or , in which case , which is impossible, as we are assuming that is nonnegative.

case 2: is negative - that is,

The equation becomes

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

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

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

Either , in which case , which is impossible since we are assuming that is negative,

or , in which case .

and can both be confirmed as solutions.

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Question

Solve for , giving all solutions, real and imaginary:

$x^{3} - 4x^{2}+ 3x - 12 = 0$

Tap to reveal answer

Answer

Factor the expression:

$x^{3} - 4x^{2}+ 3x - 12 = 0$

$\left ( x^{3} - 4x^{2} \right ) +\left ( 3x - 12 \right )= 0$

$x^{2} \left ( x - 4 \right ) + 3 \left ( x - 4 \right )= 0$

$\left (x^{2}+ 3 \right ) \left ( x - 4 \right ) = 0$

Rewrite:

or

$x^{2} +3 =0$

$x^{2} = -3$

$x = \pm \sqrt{ -3} = \pm i \sqrt{3}$

← Didn't Know|Knew It →

Question

Which of the following is the solution set of the equation:

$x-3 = \sqrt{2x+9}$

Tap to reveal answer

Answer

Square both sides of the equation, then solve the resulting quadratic equation by factoring:

$x-3 = \sqrt{2x+9}$

$\left (x-3 \right ) ^{2}=\left ( \sqrt{2x+9 \right )}^{2}$

$x^{2}-6x+9=2x+9$

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

Either

or

Checking both solutions, however:

$0-3 = \sqrt{2\cdot 0+9}$

This eliminates 0 as a solution.

$8-3 = \sqrt{2\cdot 8+9}$

8 turns out to be the only solution.

← Didn't Know|Knew It →

Question

Define $f (x) = x \cdot |x |$ . Which of the following would be a valid alternative way of expressing the definition of ?

Tap to reveal answer

Answer

By definition:

If $x \geq 0$ , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot x= x^{2}$

If , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot \left (- x \right )= - x^{2}$

← Didn't Know|Knew It →

Question

Give the solution set of the equation:

$x^{2} =|x| + 6$

Tap to reveal answer

Answer

Case 1: is nonnegative - that is, .

The equation becomes

$x^{2} =x+ 6$

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

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

Either , in which case ,

or , in which case , which is impossible, as we are assuming that is nonnegative.

case 2: is negative - that is,

The equation becomes

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

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

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

Either , in which case , which is impossible since we are assuming that is negative,

or , in which case .

and can both be confirmed as solutions.

← Didn't Know|Knew It →

Question

Solve for , giving all solutions, real and imaginary:

$x^{3} - 4x^{2}+ 3x - 12 = 0$

Tap to reveal answer

Answer

Factor the expression:

$x^{3} - 4x^{2}+ 3x - 12 = 0$

$\left ( x^{3} - 4x^{2} \right ) +\left ( 3x - 12 \right )= 0$

$x^{2} \left ( x - 4 \right ) + 3 \left ( x - 4 \right )= 0$

$\left (x^{2}+ 3 \right ) \left ( x - 4 \right ) = 0$

Rewrite:

or

$x^{2} +3 =0$

$x^{2} = -3$

$x = \pm \sqrt{ -3} = \pm i \sqrt{3}$

← Didn't Know|Knew It →

Question

Which of the following is the solution set of the equation:

$x-3 = \sqrt{2x+9}$

Tap to reveal answer

Answer

Square both sides of the equation, then solve the resulting quadratic equation by factoring:

$x-3 = \sqrt{2x+9}$

$\left (x-3 \right ) ^{2}=\left ( \sqrt{2x+9 \right )}^{2}$

$x^{2}-6x+9=2x+9$

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

Either

or

Checking both solutions, however:

$0-3 = \sqrt{2\cdot 0+9}$

This eliminates 0 as a solution.

$8-3 = \sqrt{2\cdot 8+9}$

8 turns out to be the only solution.

← Didn't Know|Knew It →

Question

Define $f (x) = x \cdot |x |$ . Which of the following would be a valid alternative way of expressing the definition of ?

Tap to reveal answer

Answer

By definition:

If $x \geq 0$ , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot x= x^{2}$

If , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot \left (- x \right )= - x^{2}$

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