GMAT Math : GMAT Quantitative Reasoning

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Example Questions

Example Question #2 : Solving Quadratic Equations

Consider the equation $x^{2} -Bx + B = 0$ . For what value(s) of does the equation have two real solutions?

Possible Answers:

B >4

B< 0 or B >4

B< -4 or B >4

B< -4

B< -4 or B >0

Correct answer:

B< 0 or B >4

Explanation:

The discriminant of the expression $x^{2} -Bx + B$ is $\left (-B \right )^{2} - 4 \cdot1\cdot B = B^{2} -4B$ . For the equation $x^{2} -Bx + B = 0$ to have two real solutions, this discriminant must be positive, so:

$B^{2} -4B > 0$

$B \left (B -4 \right ) > 0$

One of two things happens:

Case 1:

B> 0 and B -4 >0

B> 0 and B >4

But this is the same as simply saying B >4

Case 2:

B< 0 and B -4 <0

B< 0 and B <4

But this is the same as simply saying B< 0

Therefore, the equation has two real solutions if and only if either B< 0 or B >4

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Example Question #3 : Solving Quadratic Equations

Find all real solutions for :

$x^{4} + 9x ^{2}- 10 = 0$

Possible Answers:

$x \in \left \{-\sqrt{10},-1,1, \sqrt{10} \right \}$

$x \in \left \{1, \sqrt{10} \right \}$

$x \in \left \{-\sqrt{10}, \sqrt{10} \right \}$

The equation has no real solution.

$x \in \left \{ -1,1 \right \}$

Correct answer:

$x \in \left \{ -1,1 \right \}$

Explanation:

Substitute $u = x^{2}$ , and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ , and solve the resulting quadratic equation.

$x^{4} + 9x ^{2}- 10 = 0$

$u^{2} + 9u- 10 = 0$

We can rewrite the quadratic expression as (x + ?) (x + ?) , where the question marks are replaced with integers whose product is and whose sum is 9; the integers are -1,10 :

(u-1)(u+10)= 0

Set each factor to zero and solve for ; then substitute back and solve for :

u - 1 = 0

u = 1

$x^{2}= 1$

$x = -1 \textrm{ or } x = 1$

u + 10 = 0

u = -10

$x^{2}= -10$ , which has no real solution.

Therefore, the solution set is $\left \{ -1,1\right \}$

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Example Question #4 : Solving Quadratic Equations

What is the minimum value of the function $f(x) = x^{2} - 7x + 7$ for all real values of ?

Possible Answers:

$-19 \frac{1}{4}$

f(x) does not have a minimum value.

$-5 \frac{1}{4}$

$-3\frac{1}{2}$

Correct answer:

$-5 \frac{1}{4}$

Explanation:

We find the -coordinate of the vertex of the parabola for f(x) . First, we find its -coordinate using the formula

$x = -\frac{b}{2a}$ , setting a = 1, b = -7 .

$x = -\frac{b}{2a} = -\frac{-7}{2 \cdot 1} = \frac{7}{2}$

$f \left ( \frac{7}{2} \right)$ is the -coordindate of the vertex, and, subsequently, the minimum value of f(x) :

$f(x) = x^{2} - 7x + 7$

$f\left (\frac{7}{2} \right ) = \left (\frac{7}{2} \right ) ^{2} - 7\cdot \left (\frac{7}{2} \right )+ 7$

$f\left (\frac{7}{2} \right ) = \frac{49}{4} - \frac{49}{2} + 7$

$f\left (\frac{7}{2} \right ) = \frac{49}{4} - \frac{98}{4} + \frac{28}{4}$

$f\left (\frac{7}{2} \right ) = -\frac{21}{4} = -5 \frac{1}{4}$

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Example Question #5 : Solving Quadratic Equations

Define an operation as follows:

For all real numbers a,b ,

$a \; \bigstar \; b = a^{2} - ab$

Solve for : $x\; \bigstar\; 4 = 21$

Possible Answers:

$x = -1\frac{1}{4}$

$x = -3 \textrm{ or }x = 7$

x = -3

x = 7

$x = -1\frac{1}{4} \textrm{ or }x = 1\frac{1}{4}$

Correct answer:

$x = -3 \textrm{ or }x = 7$

Explanation:

Subsitute a = x, b = 4 in the defintion, and set it equal to 21:

$a \; \bigstar \; b = a^{2} - ab$

$x \; \bigstar \; 4 = 21$

$x^{2} - x \cdot 4 = 21$

$x^{2} - 4x = 21$

This sets up a quadratic equation, Move all terms to the left, factor the expression, set each factor to 0, and solve separately.

$x^{2} - 4x - 21 = 0$

(x-7)(x+3) = 0

$x-7 = 0 \Rightarrow x = 7$

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

The solution set is $\left \{ -3, 7\right \}$

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Example Question #6 : Solving Quadratic Equations

Solve for ;

(x+9)(x+4)= 28x

Possible Answers:

$x=3 \textrm{ or } x=12$

$x=-9\textrm{ or } x=-4$

$x=-12\textrm{ or } x=-3$

$x=2\textrm{ or } x=18$

$x=-18\textrm{ or } x=-2$

Correct answer:

$x=3 \textrm{ or } x=12$

Explanation:

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then colleciting all of the terms on the left side:

(x+9)(x+4)= 28x

$x^{2} + 4x + 9x + 9\cdot 4= 28x$

$x^{2} + 13x + 36= 28x$

$x^{2} + 13x + 36- 28x = 28x - 28 x$

$x^{2} -15x + 36 =0$

Now factor the quadratic expression $x^{2} -15x + 36$ to two binomial factors (x + ?)(x + ?) , replacing the question marks with two integers whose product is 36 and whose sum is . These numbers are -12,-3 , so:

(x -12)(x -3) = 0

$x-12 =0 \Rightarrow x= 12$

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

The solution set is $\left \{ 3,12 \right \}$

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Example Question #7 : Solving Quadratic Equations

Solve for :

$x ^{2} + 16 = 10x$

Possible Answers:

$x= 6 \textrm{ or }x=10$

$x= -6 \textrm{ or }x=-10$

$x= 2 \textrm{ or }x=8$

x= 4

$x= -2 \textrm{ or }x=-8$

Correct answer:

$x= 2 \textrm{ or }x=8$

Explanation:

First, rewrite the quadratic equation in standard form by moving all nonzero terms to the left:

$x ^{2} + 16 = 10x$

$x ^{2} + 16- 10x = 10x - 10x$

$x ^{2} - 10x+ 16 = 0$

Now factor the quadratic expression $x ^{2}- 10x + 16$ into two binomial factors (x + ?)(x + ?) , replacing the question marks with two integers whose product is 16 and whose sum is . These numbers are -8,-2 , so:

$x ^{2}- 10x + 16 = 0$

(x -2)(x -8) = 0

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

$x-8 = 0 \Rightarrow x=8$

The solution set is $\left \{ 2,8\right \}$ .

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Example Question #3 : Solving Quadratic Equations

Find the roots of x^2+x-2=0. Separate the answers with a comma.

Possible Answers:

2,1

-2,1

2,-1

-2,-1

Correct answer:

-2,1

Explanation:

This can be found by factoring the equation. Doing this we get

x^2+x-2=(x+2)(x-1)=0

We can solve this equation happen when x+2=0 or x-1=0. So the roots are -2,1 .

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Example Question #395 : Algebra

How many real solutions and how many imaginary solutions are there to the following quadratic equation?

$2x ^{2} + 4x = 17$

Possible Answers:

No real solutions and one imaginary solution.

Two real solutions and no imaginary solutions.

One real solution and no imaginary solutions.

No real solutions and two imaginary solutions.

One real solution and one imaginary solution.

Correct answer:

Two real solutions and no imaginary solutions.

Explanation:

Write the equation in standard form:

$2x ^{2} + 4x = 17$

$2x ^{2} + 4x - 17= 17 - 17$

$2x ^{2} + 4x - 17=0$

Evaluate the discriminant $b^{2} - 4ac$ , using a = 2, b = 4, c = -17 .

$b^{2} - 4ac = 4 ^{2} - 4 \cdot 2 \cdot (-17 ) = 16- (-136 ) = 16 + 136 = 152 > 0$

The discriminant is positive, so the equation has two distinct real solutions.

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Example Question #1471 : Problem Solving Questions

How many real solutions and how many imaginary solutions are there to the following equation?

$x^{3} -125 = 0$

Possible Answers:

One real solution, no imaginary solutions

One real solution, two imaginary solutions.

No real solutions, three imaginary solutions

No real solutions, two imaginary solutions

Three real solutions, no imaginary solutions

Correct answer:

One real solution, two imaginary solutions.

Explanation:

$x^{3} -125 =x^{3} -5^{3}$ , so this is the difference of cubes.

The difference of cubes can be factored as follows:

$A ^{3} - B^{3} = (A-B)(A^{2} +AB+ B^{2})$

Using this pattern and replacing A = x, B = 5 , this becomes:

$x ^{3} - 5^{3} = (x-5)(x^{2} +x\cdot 5+ 5^{2})$

$x ^{3} - 125 = (x-5)(x^{2} +5x+ 25)$

The equation becomes

$(x-5)(x^{2} +5x+ 25) = 0$

We set each factor equal to 0and examine the nature of the solutions.

$x - 5 = 0 \Rightarrow x = 5$ , so the equation has at least one real solution.

$x^{2} +5x+ 25 = 0$ is a quadratic equation, so we can explore these two solutions by looking at its discriminant $b^{2} - 4ac$ .

We set a = 1,b = 5, c= 25 , so the discriminant is

$b^{2} - 4ac = 5^{2} - 4\cdot 1 \cdot25= -75 < 0$

This negative disciminant indicates that there are two imaginary solutions in addition to the real one we found earlier.

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Example Question #1472 : Problem Solving Questions

For what value of does the equation $x ^{2} + 7x + C = 0$ have exactly one solution?

Possible Answers:

$C=\frac{ 49 }{4}$

C= 28

C= 14

C=49

$C=\frac{ 49 }{2}$

Correct answer:

$C=\frac{ 49 }{4}$

Explanation:

For the equation $A x^{2}+B x + C = 0$ to have exactly one solution, it must hold that the discriminant $B^{2} -4AC = 0$ , so we solve for this equation setting A = 1, B = 7 :

$B^{2} -4AC = 0$

$7^{2} -4 \cdot 1\cdot C = 0$

49 -4 C = 0

49 -4 C -49 = 0 -49

-4 C= -49

$-4 C\div (-4) = -49 \div (-4)$

$C=\frac{ 49 }{4}$

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GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #2 : Solving Quadratic Equations

Example Question #3 : Solving Quadratic Equations

Example Question #4 : Solving Quadratic Equations

Example Question #5 : Solving Quadratic Equations

Example Question #6 : Solving Quadratic Equations

Example Question #7 : Solving Quadratic Equations

Example Question #3 : Solving Quadratic Equations

Example Question #395 : Algebra

Example Question #1471 : Problem Solving Questions

Example Question #1472 : Problem Solving Questions

All GMAT Math Resources

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