SAT 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
SAT MAth and English – full syllabus practice tests
Question Medium
$
x^2-3 x-1=0
$
What is one of the solutions of the given equation?
A) \(\frac{-3+\sqrt{13}}{2}\)
B) \(\frac{-3+\sqrt{5}}{2}\)
C) \(\frac{3+\sqrt{5}}{2}\)
D) \(\frac{3+\sqrt{13}}{2}\)
▶️Answer/Explanation
Ans: D
To solve the quadratic equation \(x^2 – 3x – 1 = 0\), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
For the given equation \(x^2 – 3x – 1 = 0\), we have:
\[ a = 1, \; b = -3, \; c = -1 \]
Substitute these values into the quadratic formula:
\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}
\]
\[
x = \frac{3 \pm \sqrt{9 + 4}}{2}
\]
\[
x = \frac{3 \pm \sqrt{13}}{2}
\]
So, one of the solutions is:
\[
\boxed{\frac{3 + \sqrt{13}}{2}}
\]
Question medium
Which quadratic equation has no real solutions?
A. \(3 x^2-3=0\)
B. \(3 x^2+3 x=0\)
c. \(3 x^2+3 x+3=0\)
D. \(3 x^2-6 x+3=0\)
▶️Answer/Explanation
Ans:C
To determine which quadratic equation has no real solutions, we can use the discriminant (\(b^2 – 4ac\)) of the quadratic formula (\(x = \frac{{-b \pm \sqrt{{b^2 – 4ac}}}}{{2a}}\)). If the discriminant is negative, then the quadratic equation has no real solutions.
A. \(3x^2 – 3 = 0\)
Here, \(a = 3\), \(b = 0\), and \(c = -3\).
\[b^2 – 4ac = 0^2 – 4 \cdot 3 \cdot (-3) = 36\]
Since the discriminant is positive, this equation has real solutions.
B. \(3x^2 + 3x = 0\)
Here, \(a = 3\), \(b = 3\), and \(c = 0\).
\[b^2 – 4ac = 3^2 – 4 \cdot 3 \cdot 0 = 9\]
Since the discriminant is positive, this equation has real solutions.
C. \(3x^2 + 3x + 3 = 0\)
Here, \(a = 3\), \(b = 3\), and \(c = 3\).
\[b^2 – 4ac = 3^2 – 4 \cdot 3 \cdot 3 = -27\]
Since the discriminant is negative, this equation has no real solutions.
D. \(3x^2 – 6x + 3 = 0\)
Here, \(a = 3\), \(b = -6\), and \(c = 3\).
\[b^2 – 4ac = (-6)^2 – 4 \cdot 3 \cdot 3 = 0\]
Since the discriminant is zero, this equation has one real solution.
Question medium
An architect is asked to construct an opening in a wall in the shape of a parabola. The blueprint of the architect’s design is shown. The formula
$
y=\frac{-x(x-8)}{k}
$
where \(k\) is a constant, can be used to determine the height \(y\), in feet, of the opening at a horizontal distance of \(x\) feet from the left side of the opening. Based on the architect’s blueprint, what is the value of \(k ?\)
A. 4
B. 2
C. \(\frac{1}{2}\)
D. \(\frac{1}{4}\)
▶️Answer/Explanation
Ans:B
To find the value of \(k\), we can use the given information that when \(x = 4\), \(y = 8\). Substituting these values into the equation:
\[
8 = \frac{-4(4-8)}{k}
\]
Now, let’s solve for \(k\):
\[
8 = \frac{-4(-4)}{k}
\]
\[
8 = \frac{-16}{k}
\]
To isolate \(k\), we can cross multiply:
\[
8k = -16
\]
Now, divide both sides by 8:
\[
k = \frac{-16}{8} = -2
\]
So, the value of \(k\) is \(-2\).
Question Medium
\[
x^2+6 x+c=0
\]
In the given equation, \(c\) is a constant. The equation has exactly two distinct real solutions. Which statement about the value of \(c\) must be true?
A) \(c=6\)
B) \(c>9\)
C) \(c=9\)
D) \(c<9\)
▶️Answer/Explanation
D
To have exactly two distinct real solutions in the quadratic equation \(x^2 + 6x + c = 0\), the discriminant (\(b^2 – 4ac\)) must be positive, where \(a = 1\), \(b = 6\), and \(c\) is the constant term.
The discriminant formula is \(b^2 – 4ac\).
Substituting the given values:
\[ (6)^2 – 4(1)(c) > 0 \]
\[ 36 – 4c > 0 \]
\[ 36 > 4c \]
\[ 9 > c \]
So, the correct statement about the value of \(c\) is:
\[ \boxed{D) \, c < 9} \]
Question medium
\((3x)^{2}-4(3x)-12=0\)
What is the positive solution to the given equation?
▶️Answer/Explanation
Ans: 2
\[
(3x)^2 – 4(3x) – 12 = 0 \\
9x^2 – 12x – 12 = 0
\]
We can simplify this by dividing both sides by 3:
\[
3x^2 – 4x – 4 = 0
\]
Using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
\[
x = \frac{4 \pm \sqrt{(-4)^2 – 4(3)(-4)}}{2(3)}
\]
\[
x = \frac{4 \pm \sqrt{16 + 48}}{6}
\]
\[
x = \frac{4 \pm \sqrt{64}}{6}
\]
\[
x = \frac{4 \pm 8}{6}
\]
The positive solution is:
\[
x = \frac{4 + 8}{6} = \frac{12}{6} = 2
\]
So, the positive solution to the equation is \(x = 2\).