Solve Quadratic Equations - Praxis Core Math

1. The company breaks even when its profit is zero. A company's weekly profit $P$ from selling $x$ units is given by $P(x) = -x^2 + 50x - 225$. The company breaks even at two different production levels. What is the smaller of these two levels?
   Answer: 5 To find when the company breaks even, set the profit $P(x)$ to zero: $-x^2 + 50x - 225 = 0$. It's easier to solve if we multiply the entire equation by -1, giving $x^2 - 50x + 225 = 0$. To factor this, we need two numbers that multiply to 225 and add to -50. These numbers are -5 and -45. Thus, the equation factors to $(x-5)(x-45) = 0$. The two production levels where the company breaks even are $x=5$ and $x=45$. The smaller of these two levels is 5.
2. What positive value of $x$ satisfies the equation $2x^2 + 3x - 5 = x^2 - 2x + 19$?
   Answer: 3 First, bring all terms to one side to form a standard quadratic equation. Subtract $x^2$, add $2x$, and subtract $19$ from both sides: $(2x^2 - x^2) + (3x + 2x) + (-5 - 19) = 0$. This simplifies to $x^2 + 5x - 24 = 0$. To factor this quadratic, we need two numbers that multiply to -24 and add to 5. These numbers are 8 and -3. So the equation factors to $(x+8)(x-3) = 0$. The solutions are $x=-8$ and $x=3$. The question asks for the positive value, which is 3.
3. The height $h$, in feet, of a ball thrown upwards is given by $h = -16t^2 + 48t + 64$, where $t$ is the time in seconds. After how many seconds does the ball hit the ground?
   Answer: 4 The ball hits the ground when its height $h$ is 0. So we need to solve the equation $-16t^2 + 48t + 64 = 0$. We can simplify this equation by dividing all terms by -16: $t^2 - 3t - 4 = 0$. Factoring the quadratic expression gives $(t-4)(t+1) = 0$. The possible values for $t$ are 4 and -1. Since time cannot be negative, the ball hits the ground after 4 seconds.
4. If the roots of the equation $x^2 + bx + c = 0$ are -2 and 6, what is the value of $b$?
   Answer: -4 The sum of the roots of a quadratic equation $ax^2+bx+c=0$ is $-b/a$. In $x^2+bx+c=0$, $a=1$, so the sum of the roots is $-b$. The given roots are -2 and 6. Their sum is $-2 + 6 = 4$. Therefore, $-b = 4$, which means $b = -4$.
5. The product of two consecutive positive odd integers is 143. What is the larger of the two integers?
   Answer: 13 Let the two consecutive positive odd integers be $n$ and $n+2$. Their product is $n(n+2) = 143$. Expanding this gives $n^2 + 2n = 143$. Set the equation to zero: $n^2 + 2n - 143 = 0$. To factor this, we need two numbers that multiply to -143 and add to 2. These numbers are 13 and -11. So, $(n+13)(n-11) = 0$. The possible values for $n$ are -13 and 11. Since the integers must be positive, $n=11$. The two integers are 11 and $11+2=13$. The larger of the two is 13.
6. If $x^2 - 10x + 25 = 16$, which of the following is a possible value for $x$?
   Answer: 1 The expression on the left side, $x^2 - 10x + 25$, is a perfect square trinomial, which can be factored as $(x-5)^2$. So the equation becomes $(x-5)^2 = 16$. Take the square root of both sides: $x-5 = \pm 4$. This leads to two possibilities: $x-5 = 4$, which gives $x=9$, or $x-5 = -4$, which gives $x=1$. Of the choices provided, 1 is a possible value for $x$.
7. If $(x-3)^2 = 2x + 1$, what is the greater of the two possible values for $x$?
   Answer: $4 + 2\sqrt{2}$ First, expand the left side of the equation: $x^2 - 6x + 9 = 2x + 1$. Next, move all terms to one side to form a standard quadratic equation: $x^2 - 8x + 8 = 0$. This equation does not factor with integers, so use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here $a=1, b=-8, c=8$. So, $x = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(8)}}{2(1)} = \frac{8 \pm \sqrt{64 - 32}}{2} = \frac{8 \pm \sqrt{32}}{2}$. Simplifying the radical gives $\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}$. Thus, $x = \frac{8 \pm 4\sqrt{2}}{2} = 4 \pm 2\sqrt{2}$. The two solutions are $4 + 2\sqrt{2}$ and $4 - 2\sqrt{2}$. The greater value is $4 + 2\sqrt{2}$.
8. What is the largest real solution to the equation $x^4 - 13x^2 + 36 = 0$?
   Answer: 3 This equation is quadratic in form. Let $u = x^2$. The equation becomes $u^2 - 13u + 36 = 0$. Factoring this quadratic gives $(u-4)(u-9) = 0$. So, the solutions for $u$ are $u=4$ and $u=9$. Since $u = x^2$, we have two cases: $x^2 = 4$, which gives $x = \pm 2$, and $x^2 = 9$, which gives $x = \pm 3$. The set of all real solutions is $\{-3, -2, 2, 3\}$. The largest of these is 3.
9. Which of the following is a solution to the equation $(2x+1)^2 = 49$?
   Answer: -4 To solve the equation $(2x+1)^2 = 49$, take the square root of both sides: $2x+1 = \pm\sqrt{49}$, which means $2x+1 = \pm 7$. This gives two separate linear equations. Case 1: $2x+1 = 7$. Subtracting 1 gives $2x = 6$, so $x=3$. Case 2: $2x+1 = -7$. Subtracting 1 gives $2x = -8$, so $x=-4$. The two solutions are 3 and -4. Of the choices given, -4 is a solution.
10. The equation $3x^2 + 5x = 2$ has two real solutions. What is the value of the positive solution?
    Answer: $1/3$ First, set the equation to zero: $3x^2 + 5x - 2 = 0$. This quadratic equation can be factored into $(3x-1)(x+2) = 0$. The two solutions are found by setting each factor to zero. $3x-1=0$ gives $x = 1/3$, and $x+2=0$ gives $x = -2$. The question asks for the positive solution, which is $1/3$.
11. The lengths of the two legs of a right triangle are $x$ and $x-7$ centimeters. If the hypotenuse is 13 centimeters long, what is the value of $x$?
    Answer: 12 According to the Pythagorean theorem, $a^2 + b^2 = c^2$. For this triangle, $x^2 + (x-7)^2 = 13^2$. Expanding $(x-7)^2$ gives $x^2 - 14x + 49$. The equation becomes $x^2 + x^2 - 14x + 49 = 169$. Combining like terms and setting the equation to zero gives $2x^2 - 14x - 120 = 0$. Divide the entire equation by 2 to simplify: $x^2 - 7x - 60 = 0$. Factoring this quadratic, we need two numbers that multiply to -60 and add to -7. These are -12 and 5. So, $(x-12)(x+5) = 0$. The solutions are $x=12$ and $x=-5$. Since length must be positive, $x=12$.
12. For what positive value of $k$ does the quadratic equation $x^2 + kx + 16 = 0$ have exactly one real solution?
    Answer: 8 A quadratic equation has exactly one real solution when its discriminant is equal to zero. The discriminant is given by the expression $b^2 - 4ac$. In the equation $x^2 + kx + 16 = 0$, we have $a=1$, $b=k$, and $c=16$. Setting the discriminant to zero: $k^2 - 4(1)(16) = 0$. This simplifies to $k^2 - 64 = 0$, or $k^2 = 64$. Taking the square root of both sides gives $k = \pm 8$. Since the question asks for the positive value of $k$, the answer is 8.
13. The length of a rectangular garden is 5 feet more than its width. If the area of the garden is 84 square feet, what is the length of the garden in feet?
    Answer: 12 Let $w$ represent the width of the garden. The length is $w+5$. The area of a rectangle is length times width, so $w(w+5) = 84$. Distributing $w$ gives $w^2 + 5w = 84$. To solve this quadratic equation, set it to zero: $w^2 + 5w - 84 = 0$. Factoring the quadratic, we look for two numbers that multiply to -84 and add to 5. These numbers are 12 and -7. So, the equation factors to $(w+12)(w-7) = 0$. The possible values for $w$ are -12 and 7. Since the width cannot be negative, $w=7$ feet. The question asks for the length, which is $w+5 = 7+5 = 12$ feet.
14. If one of the solutions to $3x^2+bx-8=0$ is 2, what is the other solution?
    Answer: -4/3 Let the two solutions be $r_1$ and $r_2$. The product of the solutions is given by $c/a$. In this equation, $a=3$ and $c=-8$, so the product is $r_1 \cdot r_2 = -8/3$. We are given that one solution is $r_1 = 2$. So, $2 \cdot r_2 = -8/3$. Dividing by 2, we get $r_2 = -8/6 = -4/3$.
15. One solution to the equation $x^2 - 11x + k = 0$ is $x = 4$. What is the value of the other solution?
    Answer: 7 The sum of the two solutions of a quadratic equation $ax^2+bx+c=0$ is $-b/a$. In this equation, $a=1$ and $b=-11$, so the sum of the solutions is $-(-11)/1 = 11$. If one solution is 4, let the other solution be $r_2$. Then $4 + r_2 = 11$. Subtracting 4 from both sides gives $r_2 = 7$.
16. How many distinct real solutions does the equation $4x^2 = 3x - 1$ have?
    Answer: 0 To determine the number of distinct real solutions, first write the equation in standard form $ax^2 + bx + c = 0$: $4x^2 - 3x + 1 = 0$. Then, evaluate the discriminant, $D = b^2 - 4ac$. Here, $a=4$, $b=-3$, and $c=1$. So, $D = (-3)^2 - 4(4)(1) = 9 - 16 = -7$. Since the discriminant is negative ($D < 0$), the equation has no real solutions.
17. For which of the following values of c does the equation $2x^2 - 8x + c = 0$ have two distinct real solutions?
    Answer: 4 An equation has two distinct real solutions if its discriminant, $b^2 - 4ac$, is greater than 0. For the equation $2x^2 - 8x + c = 0$, we have $a=2$, $b=-8$, and $c$ is the unknown. The discriminant is $(-8)^2 - 4(2)(c) = 64 - 8c$. We need $64 - 8c > 0$. Adding $8c$ to both sides gives $64 > 8c$. Dividing by 8 gives $8 > c$, or $c < 8$. Of the choices provided, only 4 is less than 8.
18. A quadratic equation with integer coefficients has roots at $x = -3$ and $x = 5/2$. Which of the following could be the equation?
    Answer: $2x^2 + x - 15 = 0$ If the roots are $-3$ and $5/2$, the factors of the quadratic are $(x - (-3))$ and $(x - 5/2)$, which are $(x+3)$ and $(x - 5/2)$. The equation is $(x+3)(x - 5/2) = 0$. Expanding this gives $x^2 - \frac{5}{2}x + 3x - \frac{15}{2} = 0$. Combine the x-terms: $x^2 + \frac{1}{2}x - \frac{15}{2} = 0$. To get integer coefficients as specified, multiply the entire equation by 2: $2x^2 + x - 15 = 0$.
19. For the equation $y = (x-3)(x+5)$, for what negative value of $x$ does $y=0$?
    Answer: -5 To find the values of $x$ for which $y=0$, we set the equation to zero: $(x-3)(x+5) = 0$. By the zero product property, either $x-3=0$ or $x+5=0$. Solving these linear equations gives $x=3$ and $x=-5$. The question asks for the negative value of $x$, which is -5.
20. The equation $5x^2 + 15x - 30 = 0$ has two roots, $r_1$ and $r_2$. What is the value of $r_1 + r_2$?
    Answer: -3 The sum of the roots of a quadratic equation in the form $ax^2 + bx + c = 0$ is given by the formula $-b/a$. For the equation $5x^2 + 15x - 30 = 0$, we have $a=5$ and $b=15$. Therefore, the sum of the roots is $-15/5 = -3$.

Start this quiz