Digital SAT Math - Non-linear equations in one variable - Advanced Practice Questions - New Syllabus
DSAT MAth Practice questions – all topics
- Advanced Math Weightage: 35% Questions: 13-15
- Equivalent expressions
- Nonlinear equations in one variable and systems of equations in two variables
- Nonlinear functions
DSAT MAth and English – full syllabus practice tests
Question Hard
\[
f(x)=(2 x+3)(2 x-5)
\]
What is the minimum value of the given function?
A) -16
B) \(-\frac{3}{2}\)
C) \(\frac{1}{2}\)
D) 2
▶️Answer/Explanation
Ans: A
To find the minimum value of the function \( f(x) = (2x + 3)(2x – 5) \), we first expand it into a standard quadratic form.
\[
f(x) = (2x + 3)(2x – 5)
\]
\[
f(x) = 4x^2 – 10x + 6x – 15
\]
\[
f(x) = 4x^2 – 4x – 15
\]
The quadratic function \( f(x) = 4x^2 – 4x – 15 \) is in the standard form \( ax^2 + bx + c \), where \( a = 4 \), \( b = -4 \), and \( c = -15 \).
The minimum value of a quadratic function \( ax^2 + bx + c \) occurs at the vertex. The \(x\)-coordinate of the vertex can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Substitute \( a = 4 \) and \( b = -4 \):
\[
x = -\frac{-4}{2 \times 4} = \frac{4}{8} = \frac{1}{2}
\]
Next, substitute \( x = \frac{1}{2} \) back into the function to find the corresponding \( y \)-value:
\[
f\left( \frac{1}{2} \right) = 4 \left( \frac{1}{2} \right)^2 – 4 \left( \frac{1}{2} \right) – 15
\]
\[
f\left( \frac{1}{2} \right) = 4 \cdot \frac{1}{4} – 2 – 15
\]
\[
f\left( \frac{1}{2} \right) = 1 – 2 – 15
\]
\[
f\left( \frac{1}{2} \right) = -16
\]
Therefore, the minimum value of the function is:
\[
\boxed{-16}
\]
Question Hard
For the quadratic function \(h\), the table gives three values of \(x\) and their corresponding values of \(h(x)\). At what value of \(x\) does \(h\) reach its minimum?
A. -1
B. 0
C. 3
D. 4
▶️Answer/Explanation
Ans:C
To determine at what value of \(x\) the quadratic function \(h(x)\) reaches its minimum, we need to look for the vertex of the parabola defined by the table.
Given the values of \(h(x)\) for \(x = 2\), \(x = 4\), and \(x = 6\), we can notice that the function \(h(x)\) has a symmetric shape, typical of a quadratic function. Since the values at \(x = 2\) and \(x = 4\) are both \(0\), and the value at \(x = 6\) is \(8\), we can infer that the vertex of the parabola lies between \(x = 2\) and \(x = 6\).
The vertex of a quadratic function in the form \(h(x) = ax^2 + bx + c\) is given by the formula \(x = -\frac{b}{2a}\). However, since we don’t have the exact equation of \(h(x)\), we can still find the \(x\)-coordinate of the vertex by taking the average of the \(x\)-values where \(h(x)\) is \(0\). In this case, that would be the average of \(2\) and \(4\), which is \(3\).
Therefore, the minimum value of \(h(x)\) occurs at \(x = 3\). So, the correct answer is option C: \(3\).
Question Hard
$
x^2-4 x-9=0
$
The solutions to the given equation can be written in the form \(\frac{\mathrm{m} \pm \sqrt{\mathrm{k}}}{2}\), where \(\mathrm{m}\) and \(\mathrm{k}\) are intergers. What is the value of \(\mathrm{m}+\mathrm{k}\) ?
▶️Answer/Explanation
Ans: 56
To find the solutions to the quadratic equation \(x^2 – 4x – 9 = 0\), we can use the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\]
Where \(a = 1\), \(b = -4\), and \(c = -9\).
Substitute the values into the formula:
\[x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(-9)}}{2(1)}\]
\[x = \frac{4 \pm \sqrt{16 + 36}}{2}\]
\[x = \frac{4 \pm \sqrt{52}}{2}\]
The solutions can be written as \(m\sqrt{k}\), where \(m = 4\) and \(k = 52\). Therefore, \(m + k = 4 + 52 = 56\).
Question Hard
\[x^2+2 x-1=0\]
A solution to the given equation is \(\sqrt{k}-1\). What is the value of \(k\) ?
▶️Answer/Explanation
2
Given the equation \(x^2 + 2x – 1 = 0\) and a solution \(\sqrt{k} – 1\), let’s substitute this solution into the equation:
\[ (\sqrt{k} – 1)^2 + 2(\sqrt{k} – 1) – 1 = 0 \]
Expanding and simplifying:
\[ k – 2\sqrt{k} + 1 + 2\sqrt{k} – 2 – 1 = 0 \]
\[ k – 2 = 0 \]
So, \(k = \boxed{2}\).
Question Hard
When the quadratic function \(f\) is graphed in
the \(x y\)-plane, where \(y=f(x)\), its vertex is \((-2,5)\). One of the \(x\)-intercepts of this graph is \(\left(-\frac{7}{3}, 0\right)\). What is the other \(x\)-intercept of the graph?
A. \(\left(-\frac{13}{3}, 0\right)\)
B. \(\left(-\frac{5}{3}, 0\right)\)
C. \(\left(\frac{1}{3}, 0\right)\)
D. \(\left(\frac{7}{3}, 0\right)\)
▶️Answer/Explanation
Ans:B
Question Hard
In the xy-plane, a line with equation 2y = 4.5 intersects a parabola at exactly one point. If the parabola has equation \(y = —4x^2 + bx\), where b is a positive constant, what is the value of b?
▶️Answer/Explanation
Ans: 6
Rationale
The correct answer is 6. It’s given that a line with equation 2y = 4. 5 intersects a parabola with equation \(y = —4x^2 + bx\), where b is a positive constant, at exactly one point in the xy-plane. It follows that the system of equations consisting of 2y = 4.5 and \(y = —4x^2 + bx\) has exactly one solution. Dividing both sides of the equation of the line by 2 yields y = 2. 25. Substituting 2.25 for g in the equation of the parabola yields \(2.25 = —4x^2 + bx\). Adding \(4x^2\) and subtracting bx from both sides of this equation yields \(4^2 — bx + 2. 25 = 0\). A quadratic equation in the form of \(ax^2 + bx + c= 0\), where a, b, and c are constants, has exactly one solution when the discriminant, \(b^2 — 4ac\), is equal to zero. Substituting 4 for a and 2.25 for c in the expression \(b^2 — 4ac\) and setting this expression equal to 0 yields \(b^2 — 4(4) (2. 25) =0\), 0r b>—36=0. Adding 36 to each side of this equation yields \(b^2 = 36\). Taking the square root of each side of this equation yields b = 6. It’s given that b is positive, so the value of b is 6.
Question Hard
x—y=1
\(x+y=x^2-3\)
Which ordered pair is a solution to the system of equations above?
A \((1+\sqrt{3},\sqrt{3})\)
B.\((\sqrt{3}, -\sqrt{3})\)
C.\((1+\sqrt{5},\sqrt{5})\)
D.\((\sqrt{5}, -1+\sqrt{5})\)
▶️Answer/Explanation
Ans: A
Rationale
Choice A is correct. The solution to the given system of equations can be found by solving the first equation for x, which gives x =y +1, and substituting that value of x into the second equation which gives \(y+1+y=(y+1 )^2 —3\). Rewriting this equation by adding like terms and expanding \((y + 1)^2\) gives \(2y+1= y^2 +2y-2\). Subtracting 2y from both sides of this equation gives \(1=y^2-2\), Adding to 2 to both sides of this equation gives \(3 =y^2\). Therefore, it follows that \(y = \pm \sqrt{3}\). Substituting \(\sqrt{3}\) for y in the first equation
yields \(x —\sqrt{3} =1\). Adding \(\sqrt{3}\) to both sides of this equation yields \(x=1 +\sqrt{3}\) Therefore, the ordered pair \((1+ \sqrt{3}, \sqrt{3})\) is a solution to the given system of equations.
Choice B is incorrect. Substituting \(\sqrt{3}\) for x and \(-\sqrt{3}\) for y in the first equation yields \(\sqrt{3}—(—\sqrt{3})= 1\),0r \(2\sqrt{3}= 1\), which isn’t a true statement. Choice C is incorrect. Substituting \(1 +\sqrt{5}\) for x and \(\sqrt{5}\) for y in the second equation yields \(1 +\sqrt{5}+\sqrt{5}=(\sqrt{5})^2 —3\),0r \(1+2\sqrt{5} =2\sqrt{5} +3\), which isn’t a true statement.
Choice D is incorrect. Substituting \(\sqrt{5}\) for x and (—1 +\sqrt{5})\) for y in the second equation yields \(\sqrt{5}+(-1+\sqrt{5})=(\sqrt{5})^2-3\), which isn’t a true statement.
Question Hard
\(\frac{1}{x^2+10x+25}=4\)
If x is a solution to the given equation, which of the following is a possible value of x +5 ?
A.\(\frac{1}{2}\)
B.\(\frac{5}{2}\)
C.\(\frac{9}{2}\)
D.\(\frac{11}{2}\)
▶️Answer/Explanation
Ans: A
Rationale
Choice A is correct. The given equation can be rewritten as \(\frac{1}{(x+5)^2}=4\). Multiplying both sides of this equation by \((x+5)^2\) yields \(1= 4(x +5)^2\) . Dividing both sides of this equation by 4 yields \(\frac{1}{4}=(x+5)^2\). Taking the square root of both sides of this equation yields \(\frac{1}{2}=(x+5)\) or \(-\frac{1}{2}=(x+5)\) . Therefore, a possible value of x+5 is \(\frac{1}{2}\).
Choices B, C, and D are incorrect and may result from computational or conceptual errors.
Question Hard
During a 5-second time interval, the average acceleration a, in meters per second squared, of an object with an initial velocity of 12 meters per second is defined by the equation \(a=\frac{v_f-12}{5}\), where \(v_f\) is the final velocity of the object in meters per second. If the equation is rewritten in the form \(v_f =xa +y\), where x and y are constants, what is the value of x?
▶️Answer/Explanation
Ans: Rationale
The correct answer is 5. The given equation can be rewritten in the form \(v_f =xa +y\) like so:
\(a=\frac{v_f-12}{5}\)
\(v_f—12=5a\)
\(v_f=5a+12\)
It follows that the value of x is 5 and the value of y is 12.
Question Hard
\(2x^2—2=2x+3\)
Which of the following is a solution to the equation above?
A.2
B.\(1-\sqrt{11}\)
C.\(\frac{1}{2}+\sqrt{11}\)
D.\(\frac{1+\sqrt{11}}{2}\)
▶️Answer/Explanation
Ans: D
Rationale
Choice D is correct. A quadratic equation in the form \(ax^2 +bx+c= 0\), where a, b, and c are constants, can be solved using the quadratic formula: \(x=-\frac{(-b)\pm \sqrt{b^2-4ac}}{2a}\). Subtracting 2x + 3 from both sides of the given equation yields \(2x^2-2x—5= 0\). Applying the quadratic formula, where a =2, b = —2, and c = —5, yields \(x=-\frac{(-(-2))\pm \sqrt{(-2)^2-4.2.-5}}{2.2}\). This can be rewritten as \(x=\frac{2\pm\sqrt{44}}{4}\). Since, \(\sqrt{44}=\sqrt{2^2(11)}\) or \(2\sqrt{11}\), the equation can be rewritten as \(x=\frac{2\pm 2 \sqrt{11}}{4}\). Dividing 2 from both the numerator and denominator yields \(\frac{2- 2 \sqrt{11}}{4}\) or \(\frac{2+2 \sqrt{11}}{4}\). Of these two solutions, only \(\frac{2+2 \sqrt{11}}{4}\) is present among the choices. Thus, the correct choice is D.
Choice A is incorrect and may result from a computational or conceptual error. Choice B is incorrect and may result from using \(x=-\frac{(-b)\pm \sqrt{b^2-4ac}}{a}\) instead of \(x=-\frac{(-b)\pm \sqrt{b^2-4ac}}{2a}\) as the quadratic formula. Choice C is incorrect and may result from rewriting \(\sqrt{44}\) instead of \(2\sqrt{11}\).
Question Hard
If \(3x^2 — 18x— 15 = 0\), what is the value of \(x^2 — 6x\)?
▶️Answer/Explanation
Ans: 5
Rationale
The correct answer is 5. Dividing each side of the given equation by 3 yields \(x^2 — 6x — 5 = 0\). Adding 5 to each side of this equation yields \(x^2 — 6x = 5\). Therefore, if \(3x^2 — 18x — 15 = 0\), the value of \(x^2 — 6x\) is 5.
Question Hard
\(2x^2 —4x =t\)
In the equation above, t is a constant. If the equation has no real solutions, which of the following could be the value of t ?
A -3
B.—1
C.1
D.3
▶️Answer/Explanation
Ans: A
Rationale
Choice A is correct. The number of solutions to any quadratic equation in the form \(ax^2+bx+c= 0\), where a, b, and c are constants, can be found by evaluating the expression b2 —4ac, which is called the discriminant. If the value of \(b^2—4ac\) is a positive number, then there will be exactly two real solutions to the equation. If the value of \(b^2—4ac\) is zero, then there will be exactly one real solution to the equation. Finally, if the value of \(b^2 —4ac\) is negative, then there will be no real solutions to the equation.
The given equation \(2x^2—4x=t\) is a quadratic equation in one variable, where t is a constant. Subtracting t from both sides of the equation gives \(2x^2—4x—t=0\). In this form, a=2, b = —4,and c = —t. The values of t for which the equation has no real solutions are the same values of t for which the discriminant of this equation is a negative value. The discriminant is equal to \((-4)^2 -4(2)(-t)\); therefore, \((-4)^2 -4(2)(-t)< 0\). Simplifying the left side of the inequality gives 16 + 8t < 0. Subtracting 16 from both sides of the inequality and then dividing both sides by 8 gives t < —2. Of the values given in the options, —3 is the only value that is less than —2. Therefore, choice A must be the correct answer.
Choices B, C, and D are incorrect and may result from a misconception about how to use the discriminant to determine the number of solutions of a quadratic equation in one variable.
Question Hard
\(\sqrt{2x+6}+4=x+3\)
What is the solution set of the equation above?
A {1}
B. {5}
C.{-1, 5}
D.{0, —-1, 5}
▶️Answer/Explanation
Ans: B
Rationale
Choice B is correct. Subtracting 4 from both sides of \(\sqrt{2x+6}+4=x+3\) isolates the radical expression on the left side of the equation as follows: \(\sqrt{2x+6}=x-1\). Squaring both sides of \(\sqrt{2x+6}=x-1\) yields \(2x+6=x^2-2x+1\). This equation can be rewritten as a quadratic equation in standard form: \(x^2—4x—-5= 0\). One way to solve this quadratic equation is to factor the expression \(x^2—4x-5\) by identifying two numbers with a sum of —4 and a product of —5. These numbers are —5 and 1. So the quadratic equation can be factored as (x—5)(x+1)=0. It follows that 5 and —1 are the solutions to the quadratic equation. However, the solutions must be verified by checking whether 5 and—1 satisfy the original equation, \(\sqrt{2x+6}+4=x+3\).
When x = — 1, the original equation gives \(\sqrt{2(-1)+6}+4=(-1)+3\), or 6 =2, which is false. Therefore, —1 does not satisfy the original equation. When x = 5, the original equation gives \(\sqrt{2(5)+6}+4=5+3\),0r 8=8, which is true. Therefore, x =5 is the only solution to the original equation, and so the solution set is {5}.
Choices A, C, and D are incorrect because each of these sets contains at least one value that results in a false statement when substituted into the given equation. For instance, in choice D, when 0 is substituted for x into the given equation, the result is \(\sqrt{2(0)+6}+4=0+3\), or \(\sqrt{6}+4=3\). This is not a true statement, so 0 is not a solution to the given equation.
Question Hard
The graph of the function f, defined by \(f(x)=-\frac{1}{2}(x-4)^2+10\), is shown in the xy-plane above. If the function g (not shown) is defined by g(x) = —x+10, what is one possible value of a such that f(a)=g(a) ?
▶️Answer/Explanation
Ans: Rationale
The correct answer is either 2 or 8. Substituting x=a in the definitions for f and g gives \(f(a)=-\frac{1}{2}(a-4)^2+10\) and g(a) = — a+10, respectively. If f(a)=g(a), then \(-\frac{1}{2}(a-4)^2+10=— a+10\). Subtracting 10 from both sides of this equation gives \(-\frac{1}{2}(a-4)^2=— a\). Multiplying both sides by —2 gives \((a— 4)^2 =2a\). Expanding \((a— 4)^2\) gives \(a^2-8a+16= 2a\). Combining the like terms on one side of the equation gives \(a^2-10a+16= 0\). One way to solve this equation is to factor \(a^2-10a+16\) by identifying two
numbers with a sum of -10 and a product of 16. These numbers are —2 and —8, so the quadratic equation can be factored as (a—2)(a—8) = 0. Therefore, the possible values of a are either 2 or 8. Note that 2 and 8 are examples of ways to enter a correct answer.
Alternate approach: Graphically, the condition f(a) = g(a) implies the graphs of the functions y =f(x) and y= g(x) intersect at x=a. The graph y = f(x) is given, and the graph of y= g(x) may be sketched as a line with y-intercept 10 and a slope of —1 (taking care to note the different scales on each axis). These two graphs intersect at x =2 and x = 8.
Question Hard
\(—16x^2 — 8x +c=0\)
In the given equation, c is a constant. The equation has exactly one solution. What is the value of c?
▶️Answer/Explanation
Ans: -1
Question Hard
\((x—1)^2=—4\)
How many distinct real solutions does the given equation have?
A. Exactly one
B. Exactly two
C. Infinitely many
D. Zero
▶️Answer/Explanation
Ans: D
There is no intersection point. Hence, the system of equation has zero solution.
Question Hard
\(y=x^2+2x+1\)
x+y+1=0
If \((x_1,y_1)\) and \((x_2,y_2)\) are the two solutions to the system of equations above, what is the value of \(y_1+y_2\)?
A -3
B.-2
C.-1
D. 1
▶️Answer/Explanation
Ans: D
The two intersection points are (-2,1) and (-1,0).
\(y_1+y_2=1+0=1\)
Question Hard
If \(u-3=\frac{6}{t-2}\) what is t in terms of u?
A.\(t=\frac{1}{u}\)
B.\(t=\frac{2u+9}{u}\)
C.\(t=\frac{1}{u-3}\)
D.\(t=\frac{2u}{u-3}\)
▶️Answer/Explanation
Ans: D
Multiply and Divide by \(\frac{t-2}{u-3}\) on both sides, we get
\(t-2=\frac{6}{u-3}\)
\(t=\frac{2u-6+6}{u-3}\)
\(t=\frac{2u}{u-3}\)
Question Hard
In the xy-plane, the graph of \(y =x^2-9\) intersects line p at (1 ,a) and (5,b), where a and b are constants. What is the slope of line p ?
A 6
B. 2
C.-2
D. -6
▶️Answer/Explanation
Ans: A
Question Hard
\(D=T-\frac{9}{25}(100-H)\)
The formula above can be used to approximate the dew point D, in degrees Fahrenheit, given the temperature T, in degrees Fahrenheit, and the relative humidity of H percent, where H > 50. Which of the following expresses the relative humidity in terms of the temperature and the dew point?
A.\(H=\frac{25}{9}(D-T)+100\)
B.\(H=\frac{25}{9}(D-T)-100\)
C.\(H=\frac{25}{9}(D+T)+100\)
D.\(H=\frac{25}{9}(D+T)-100\)
▶️Answer/Explanation
Ans: A
Subtract T from both sides, we get
\(D-T=-\frac{9}{25}(100-H)\)
Multiply by \(-\frac{25}{9}\) on both sides, we get
\(-\frac{25}{9}(D-T)=100-H\)
Solve for H.
\(H=\frac{25}{9}(D-T)+100\)
Question Hard
In the xy-plane, the graph of \(y= 3x^2— 14x\) intersects the graph of y=x at the points (0,0) and (a, a). What is the value of a?
▶️Answer/Explanation
Ans: 5
Question Hard
\(x^2— 40x -10=0\)
What is the sum of the solutions to the given equation?
A.0
B.5
C.10
D. 40
▶️Answer/Explanation
Ans: D
The sum of the solution is \(-\frac{b}{a}\).
Here, \(b=-40\) and \(a=1\).
The sum of the solution is \(-\frac{-40}{1}=40\).
Question Hard
\(y=x^2+3x-7\)
y—5x+8=0
How many solutions are there to the system of equations above?
A. There are exactly 4 solutions.
B. There are exactly 2 solutions.
C. There is exactly 1 solution.
D. There are no solutions.
▶️Answer/Explanation
Ans: C
Question Hard
\(\frac{4x^2}{x^2-9}-\frac{2x}{x+3}=\frac{1}{x-3}\)
What value of x satisfies the
equation above?
A. -3
B.\(-\frac{1}{2}\)
C.\(\frac{1}{2}\)
D.3
▶️Answer/Explanation
Ans: C
Question Hard
\(x^2 +y+7=7\)
\(20x+100—y=0\)
The solution to the given system of equations is (x, y). What is the value of x?
▶️Answer/Explanation
Ans: -10
Solve equation 2 for \(y\).
\(y=20x+100\)
Replace \(y\) with \(20x+100\).
\(x^2+20x+100+7=7\)
Question Hard
5(x+7) = 15(x — 17)(x + 7) What is the sum of the solutions to the given equation?
▶️Answer/Explanation
Ans: 10.33,31/3
\(5(x+7) – 15(x — 17)(x + 7)=0\)
\(5(x+7)(1-3(x-17))=0\)
\(x_1=-7\) or \(x_2=\frac{1}{3}+17\)
Sum of the solution: \(-7+\frac{1}{3}+17=\frac{31}{3}\) or \(10.33\)
Question Hard
The solutions to \(x^2 + 6x + 7 = 0\) are r and s, where r < s. The solutions to \(x^2 + 8x + 8 = 0\) are t and u, where t < u. The solutions to \(x^2 + 14x + c= 0\), where c is a constant, are r+t and s + u. What is the value of c?
▶️Answer/Explanation
Ans: 31
Given the equations:
1. \( x^2 + 6x + 7 = 0 \) → roots: \( r, s \)
2. \( x^2 + 8x + 8 = 0 \) → roots: \( t, u \)
3. \( x^2 + 14x + c = 0 \) → roots: \( r+t, s+u \)
Use the product of roots formula:
$
rs = 7, \quad tu = 8
$
$
(r+t)(s+u) = rs + ru + st + tu
$
Use sum-product relations:
$
ru + st = (r+s)(t+u) – rs – tu
$
Since \( r+s = -6 \) and \( t+u = -8 \):
$
ru + st = (-6)(-8) – 7 – 8 = 48 – 15 = 33
$
$
c = rs + ru + st + tu = 7 + 33 + 8 = 31
$
Question Hard
1. Which expression is equivalent to \( (x^2 y)(x^4 y^{-3}) \), where \( x, y, \) and \( z \) are positive numbers?
A) \( x^6 y^{-3} \)
B) \( x^6 y^{-2} \)
C) \( x^4 y^{-3} \)
D) \( x^2 y^{-2} \)
▶️Answer/Explanation
Answer: B
Use exponent rules:
When multiplying terms with the same base, add the exponents.
Simplify \( x \)-terms:
$
x^2 \cdot x^4 = x^{2+4} = x^6
$
Simplify \( y \)-terms:
$
y \cdot y^{-3} = y^{1+(-3)} = y^{-2}
$
Final expression:
$
x^6 y^{-2}
$
Question Hard
2.Which of the following is equivalent to \( \sqrt{\frac{x}{64}} \) for all \( x > 0 \)?
A) \( \frac{x^2}{8} \)
B) \( \frac{x^2}{32} \)
C) \( \frac{x^{\frac{1}{2}}}{8} \)
D) \( \frac{x^{\frac{1}{2}}}{32} \)
▶️Answer/Explanation
Answer: C
Simplify \( \sqrt{\frac{x}{64}} \):
$
\sqrt{\frac{x}{64}} = \frac{\sqrt{x}}{\sqrt{64}} = \frac{x^{\frac{1}{2}}}{8}
$
Question Hard
3. Which expression is equivalent to \( \frac{x^{\frac{5}{2}}}{\sqrt{x}} \), where \( x \neq 0 \)?
A) \( x^{\frac{3}{2}} \)
B) \( x^{\frac{5}{3}} \)
C) \( x^{\frac{3}{2}} \)
D) \( x^2 \)
▶️Answer/Explanation
Answer: D
Simplify:
$
\frac{x^{\frac{5}{2}}}{\sqrt{x}} = \frac{x^{\frac{5}{2}}}{x^{\frac{1}{2}}}
$
Apply exponent rule (\( a^m / a^n = a^{m-n} \)):
$
x^{\frac{5}{2} – \frac{1}{2}} = x^{\frac{4}{2}} = x^2
$
Question Hard
4. Which expression is equivalent to \( b^{\frac{y}{x}} \), where \( b > 0 \)?
A) \( \sqrt[5]{b^{105}} \)
B) \( \sqrt[5]{b^{135}} \)
C) \( \sqrt[105]{b^{135}} \)
D) \( \sqrt[135]{b^{105}} \)
▶️Answer/Explanation
Answer: C
Convert to Radical Form:
$
b^{\frac{y}{x}} = \sqrt[x]{b^y}
$
The correct form is \( \sqrt[105]{b^{135}} \), since \( \frac{135}{105} = \frac{y}{x} \).
Question Hard
5. Which expression represents the product of \( (b^6 c^{-2} d^{-5}) \) and \( (b^8 c^{-3} + c^4 d^5) \), where \( b, c, \) and \( d \) are positive?
A) \( b^{14} c^{-5} + c^2 d^{-10} \)
B) \( b^{14} c^{-5} + c^2 \)
C) \( b^{14} c^{-5} d^{-5} + b^6 c^2 \)
D) \( b^{14} c^{-5} d^{-5} + c^2 \)
▶️Answer/Explanation
Answer: C
Multiply the Terms:
$
(b^6 c^{-2} d^{-5}) \cdot (b^8 c^{-3} + c^4 d^5)
$
Distribute:
$
b^6 c^{-2} d^{-5} \cdot b^8 c^{-3} + b^6 c^{-2} d^{-5} \cdot c^4 d^5
$
Simplify Each Term:
1. First term:
$
b^{6+8} c^{-2-3} d^{-5} = b^{14} c^{-5} d^{-5}
$
2. Second term:
$
b^6 c^{-2+4} d^{-5+5} = b^6 c^2 d^0 = b^6 c^2
$
Question Hard
6. If \( x \neq 0 \) and \( y \neq 0 \), which of the following is equivalent to \( \frac{8x^2}{\sqrt{4x^6 y^4}} \)?
A) \( 2xy^{-2} \)
B) \( 4x^{-2} y^2 \)
C) \( 4x^{-1} y^{-2} \)
D) \( 4xy^2 \)
▶️Answer/Explanation
Answer:
Question Hard
7. If \( r \) and \( s \) are positive, which of the following expressions is equivalent to \( \frac{r^3}{s^7} \)?
A) \( \frac{1}{\sqrt[7]{r^5 s^{14}}} \)
B) \( \sqrt[7]{r^5 s^{14}} \)
C) \( \frac{1}{\sqrt[7]{r^6 s^3}} \)
D) \( \sqrt[7]{r^9 s^3} \)
▶️Answer/Explanation
Answer:
Question Hard
8. The given equation \( \sqrt[d]{a} = \sqrt[b]{c} \) relates the distinct positive numbers \( a, b, c, \) and \( d \). Which equation correctly expresses \( c \) in terms of \( a, b, \) and \( d \)?
A) \( c = \frac{d}{a^{\frac{1}{b}}} \)
B) \( c = a^{d – h} \)
C) \( c = a^{\frac{d}{b}} \)
D) \( c = a^{\frac{b}{d}} \)
▶️Answer/Explanation
Answer:
Question Hard
9. Which expression is equivalent to \( \sqrt{16a^3} \), where \( a > 0 \)?
A) \( 4a^{\frac{3}{2}} \)
B) \( 4a^{\frac{5}{3}} \)
C) \( 8a^{\frac{3}{4}} \)
D) \( 8a^{\frac{4}{3}} \)
▶️Answer/Explanation
Answer:
Question Hard
10. Which expression is equivalent to \( \sqrt{r s} (\sqrt{r} + \sqrt{s}) \), where \( r \geq 0 \) and \( s \geq 0 \)?
A) \( \sqrt{r^2 s} + r s^{\frac{5}{2}} \)
B) \( r \sqrt{s} + s \sqrt{r} \)
C) \( rs \sqrt{r} + s \)
D) \( \sqrt{rs} + r + s \)
▶️Answer/Explanation
Answer:
Question Hard
11. The expression \( \left( \sqrt{x^2} \right)^n \), where \( n \) is a constant, is equivalent to \( x^8 \). What is the value of \( n \)?
▶️Answer/Explanation
Answer:
Question Hard
12. Which expression is equivalent to \( 2^{-2k} \cdot 3^k \)?
A) \( (3\sqrt{2})^k \)
B) \( \left( \frac{1}{36} \right)^k \)
C) \( \left( \frac{3}{4} \right)^k \)
D) \( \left( \frac{9}{4} \right)^k \)
▶️Answer/Explanation
Answer:
Question Hard
13. Which expression is equivalent to \( a^{\frac{2}{6}} \cdot (a^{\frac{5}{3}})^{\frac{1}{2}} \), where \( a \) is positive?
A) \( \sqrt[3]{a^2} \)
B) \( \sqrt[3]{a^{16}} \)
C) \( \sqrt[3]{a^8} \)
D) \( \sqrt[3]{a^{14}} \)
▶️Answer/Explanation
Answer:
Question Hard
\[
\frac{(x-4)(x+2)}{(x-4)}=0
\]
Which value is a solution to the given equation?
A) 4
B) 2
C) 0
D) -2
▶️Answer/Explanation
D
Given the equation:
\[ \frac{(x-4)(x+2)}{(x-4)} = 0 \]
First, simplify the equation:
\[ x + 2 = 0 \]
\[ x = -2 \]
We must also consider the domain restriction from the denominator \( (x-4) \):
\[ x \neq 4 \]
Since \( x = -2 \) does not violate this restriction, it is a valid solution.
So the answer is:
\[ \boxed{D} \]
Question Hard
The table shows several values of x and their corresponding values of f(x). The function f is defined by f(x) = mx + b, where m and b are constants. What is the value of b ?
▶️Answer/Explanation
Ans: 16
To find the value of \(b\), we can use the fact that \(f(x) = mx + b\).
Let’s choose the point \((2, 106)\).
Substituting \(x = 2\) and \(f(x) = 106\) into the equation, we get:
\[106 = m \cdot 2 + b\]
Now, we need to find the slope \(m\) of the function. use another point from the table, say \((3, 151)\), to find \(m\).
Substituting \(x = 3\) and \(f(x) = 151\) into the equation, we get:
\[151 = m \cdot 3 + b\]
Now, we have two equations:
\[106 = 2m + b\]
\[151 = 3m + b\]
We can subtract the first equation from the second to eliminate \(b\):
\[151 – 106 = 3m – 2m\]
\[45 = m\]
Now, we can substitute \(m = 45\) into one of the equations to solve for \(b\).
\[106 = 2 \cdot 45 + b\]
\[106 = 90 + b\]
\[b = 106 – 90\]
\[b = 16\]
So, the value of \(b\) is \(\mathbf{16}\).
Question Hard
$
|x+1|=5
$
What positive value of \(\mathrm{x}\) satisfies the given equation?
▶️Answer/Explanation
Ans:4
To solve \(|x + 1| = 5\), we consider the definition of absolute value. This gives us two equations:
\[x + 1 = 5\]
\[x + 1 = -5\]
Solving these:
\[x + 1 = 5 \implies x = 4\]
\[x + 1 = -5 \implies x = -6\]
The positive value of \(x\) that satisfies the equation is:
\[x = 4\]
Question Hard
$$
(x-3)^4=0
$$
What value of $x$ makes the equation above true?
▶️Answer/Explanation
3