Reviewed by Janaisa Harris
Janaisa Harris, BA-Mathematics |
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Ms. Janaisa Harris, an experienced educator, has devoted 4 years to teaching high school math and 6 years to tutoring. She holds a degree in Mathematics (Secondary Education, and Teaching) from the University of North Carolina at Greensboro and is currently employed at Wilson County School (NC) as a mathematics teacher. She is now broadening her educational impact by engaging in curriculum mapping for her county. This endeavor enriches her understanding of educational strategies and their implementation. With a strong commitment to quality education, she actively participates in the review process of educational quizzes, ensuring accuracy and relevance to the curriculum.
, BA-Mathematics
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Take our " Quadratic Equations Practice Test Questions and Answers " to check your knowledge on this topic. Quadratic equations are an important topic in mathematics. All the students need to learn and should have a good command of this important topic. In this quiz, you just have to pick the correct option from the other option choices given below to get a great score. Additionally, this quiz is also good if you want to prepare for your quadratic test. We are sure that you Have fun learning Maths!

• 1.

### -1x2 + 0x + 49 = 0

• A.

X = -9 and -6

• B.

X = 7 and -7

• C.

X = 8 and 3

• D.

X = 7 and -3

• E.

X = 9 and -9

B. X = 7 and -7
Explanation
The given equation is a quadratic equation of the form ax^2 + bx + c = 0. By factoring or using the quadratic formula, we can find the values of x that satisfy the equation. In this case, the equation can be factored as (x - 7)(x + 7) = 0, which means that x = 7 and x = -7 are the solutions to the equation.

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• 2.

### -1x2 + 2x + 48 = 0

• A.

X = -2 and 1

• B.

X = -1 and -7

• C.

X = 9 and -9

• D.

X = 8 and -8

• E.

X = 8 and -6

E. X = 8 and -6
Explanation
Sure, I’d be happy to explain how to solve the equation
-1x^2 + 2x + 48 = 0

This is a quadratic equation, and it can be solved using the quadratic formula, which is
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this equation, a is the coefficient of x^2, b is the coefficient of x, and c is the constant term. So, for our equation, a is -1, b is 2, and c is 48.

Substituting these values into the quadratic formula gives us:
x = \frac{-2 \pm \sqrt{(2)^2 - 4*(-1)*48}}{2*(-1)}
x = \frac{-2 \pm \sqrt{4 + 192}}{-2}
x = \frac{-2 \pm \sqrt{196}}{-2}
x = \frac{-2 \pm 14}{-2}

This gives us two solutions:
x = \frac{-2 - 14}{-2} = 8
x = \frac{-2 + 14}{-2} = -6

So, the solutions to the equation
-1x^2 + 2x + 48 = 0
are x = 8 and x = -6.

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• 3.

### 1x2 + 5x - 14 = 0

• A.

X = -1 and 2

• B.

X = -1 and 2

• C.

X = -7 and 2

• D.

X = 9 and -9

• E.

X = 9 and -9

C. X = -7 and 2
Explanation
The given equation is a quadratic equation in the form of ax^2 + bx + c = 0. By factoring or using the quadratic formula, we can find the values of x that satisfy the equation. In this case, the equation can be factored as (x - 2)(x + 7) = 0, which means that x can be either -7 or 2. Therefore, the correct answer is x = -7 and 2.

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• 4.

### 1x2 + 10x + 21 = 0

• A.

X = -7 and -3

• B.

X = -7 and 3

• C.

X = 8 and -6

• D.

X = 8 and 6

• E.

X = 10 and 11

A. X = -7 and -3
Explanation
The given equation is a quadratic equation in the form of ax^2 + bx + c = 0. By factoring or using the quadratic formula, we can find the values of x that satisfy the equation. In this case, the correct answer is x = -7 and -3 because when we substitute these values into the equation, we get 1*(-7)^2 + 10*(-7) + 21 = 0 and 1*(-3)^2 + 10*(-3) + 21 = 0, which are both true.

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• 5.

### -1x2 + 3x + 28 = 0

• A.

X = -6 and -8

• B.

X = 9 and 4

• C.

X = 6 and -5

• D.

X = -7 and -4

• E.

X = 7 and 4

E. X = 7 and 4
Explanation
The given equation is a quadratic equation in the form of ax^2 + bx + c = 0.
Using the formula, x = (-b ± √(b^2 - 4ac)) / 2a
Given the coefficients of the equation: a = -1 b = 3 c = 28
Plugging these values into the formula, we get: x = (-(3) ± √((3)^2 - 4(-1)(28))) / 2(-1) x = (-3 ± √(9 + 112)) / -2 x = (-3 ± √121) / -2 x = (-3 ± 11) / -2
So, the solutions for various values of x are:
x = (11 - 3) / -2 = 4
x = (-3 - 11) / -2 = 7

By solving the equation, we find that the values of x that satisfy the equation are 7 and 4. These values are the solutions to the equation and make it true when substituted into the equation. Therefore, the correct answer is x = 7 and 4.

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• 6.

### What is the vertex of the following equation: x2 - 8x + 15 = 0?

• A.

(4,1)

• B.

(4,-1)

• C.

(-4,-1)

• D.

(-4,1)

B. (4,-1)
Explanation
The vertex of a quadratic equation in the form of y = ax^2 + bx + c is given by the coordinates (-b/2a, f(-b/2a)). In this equation, a = 1, b = -8, and c = 15. Plugging these values into the formula, we get (-(-8)/2(1), f(-(-8)/2(1))). Simplifying further, we get (4, f(4)). To find the y-coordinate, we substitute x = 4 into the equation: 4^2 - 8(4) + 15 = 1. Therefore, the vertex is (4, -1).

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• 7.

### What is the  axis of symmetry and range of the following function: x2 - 8x + 15 = 0?

• A.

Axis: x=4 ; Range: (-1,infinity)

• B.

Axis: x=-4 ; Range: (-1, infinity)

• C.

Axis: x=-1 ; Range: (4, infinity)

• D.

Axis: x=-1 ; Range: (-4, infinity)

A. Axis: x=4 ; Range: (-1,infinity)
Explanation
The given quadratic function is in the form of f(x) = x^2 - 8x + 15. To find the axis of symmetry, we can use the formula x = -b/2a, where a, b, and c are coefficients of the quadratic equation. In this case, a = 1 and b = -8. Substituting these values into the formula, we get x = -(-8)/2(1), which simplifies to x = 4. Therefore, the correct answer is "Axis: x=4".

To find the range of the function, we need to determine the set of all possible y-values that the function can produce. Since the coefficient of x^2 is positive, the parabola opens upward and the minimum value occurs at the vertex. The y-coordinate of the vertex can be found by substituting the x-coordinate of the axis of symmetry (4) into the function. f(4) = (4)^2 - 8(4) + 15 = 16 - 32 + 15 = -1. Hence, the range of the function is (-1, infinity). Therefore, the correct answer is "Range: (-1, infinity)".

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• 8.

### What is the vertex of the following equation:  -x2 - 9x - 8 = 0?

• A.

(-4.5, 12.25)

• B.

(-1,-4.5)

• C.

(-1,9)

• D.

(1,9)

A. (-4.5, 12.25)
Explanation
The vertex of a quadratic equation in the form of y = ax^2 + bx + c can be found using the formula x = -b/2a. In this equation, a = -1 and b = -9. Plugging these values into the formula, we get x = -(-9)/2(-1) = 9/-2 = -4.5. To find the y-coordinate of the vertex, we substitute this value of x back into the equation: y = -(-4.5)^2 - 9(-4.5) - 8 = -20.25 + 40.5 - 8 = 12.25. Therefore, the vertex is (-4.5, 12.25).

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• 9.

### What is the range of the following function:  -x2 + 2x + 8 = 0?

• A.

(infinity, 9)

• B.

(-infinity, infinity)

• C.

(9, infinity)

• D.

(-9, infinity)

• E.

(-infinity, 9)

E. (-infinity, 9)
Explanation
For the given quadratic function  -x2 + 2x + 8 = 0, it forms a downward-opening parabola. This means that the highest point of the parabola, known as the vertex, represents the maximum value of the function.
By calculating the vertex of the parabola, we find that it occurs at the point (1, 9), where the x-coordinate is 1 and the y-coordinate (or function value) is 9.
Since the parabola opens downwards, the maximum value of the function is 9. Therefore, the range of the function consists of all real numbers less than or equal to 9. In simpler terms, the function's range includes all values from negative infinity up to and including 9.

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• 10.

### What is the domain of the following function:  -x2 + 2x + 8 = 0?

• A.

(1,9)

• B.

(-infinity,infinity)

• C.

(infinity,9)

• D.

(1, infinity)

• E.

(9,infinity)

B. (-infinity,infinity)
Explanation
The function -x^2 + 2x + 8 = 0 is a quadratic equation. The domain of a quadratic equation is always the set of all real numbers, which means that the function is defined for any value of x. Therefore, the correct answer is (-infinity, infinity).

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Janaisa Harris |BA-Mathematics |
Mathematics Expert
Ms. Janaisa Harris, an experienced educator, has devoted 4 years to teaching high school math and 6 years to tutoring. She holds a degree in Mathematics (Secondary Education, and Teaching) from the University of North Carolina at Greensboro and is currently employed at Wilson County School (NC) as a mathematics teacher. She is now broadening her educational impact by engaging in curriculum mapping for her county. This endeavor enriches her understanding of educational strategies and their implementation. With a strong commitment to quality education, she actively participates in the review process of educational quizzes, ensuring accuracy and relevance to the curriculum.

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