# Digital SAT Math Practice Questions – Medium : Solving quadratic equations

## 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

[calc]  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) $$\frac{3+\sqrt{13}}{2}$$

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:
$\boxed{\frac{3 + \sqrt{13}}{2}}$

[Calc]  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. $$3 x^2-6 x+3=0$$

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.

[No calc]  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. $$\frac{1}{4}$$

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 $$-2$$.

[Calc]  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) $$c<9$$

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:
$\boxed{D) \, c < 9}$

[No calc]  Question   medium

$$(3x)^{2}-4(3x)-12=0$$

What is the positive solution to the given equation?

Ans: 2

$(3x)^2 – 4(3x) – 12 = 0 \\ 9x^2 – 12x – 12 = 0$

We can simplify this by dividing both sides by 3:

$3x^2 – 4x – 4 = 0$

$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

$x = \frac{4 \pm \sqrt{(-4)^2 – 4(3)(-4)}}{2(3)}$

$x = \frac{4 \pm \sqrt{16 + 48}}{6}$

$x = \frac{4 \pm \sqrt{64}}{6}$

$x = \frac{4 \pm 8}{6}$

The positive solution is:

$x = \frac{4 + 8}{6} = \frac{12}{6} = 2$

So, the positive solution to the equation is $$x = 2$$.

[Calc]  Question  medium

Which quadratic equation has exactly one
distinct real solution?
A. $$x^{2}+4x+16=0$$
B. $$x^{2}-8x-16=0$$
C.  $$x^{2}-6x-16=0$$
D. $$x^{2}-8x+16=0$$

Ans: D

A quadratic equation has exactly one distinct real solution when its discriminant ($$b^2 – 4ac$$) is equal to zero.

Let’s check the discriminants for each equation:

A. $$x^2 + 4x + 16 = 0$$
Discriminant: $$4^2 – 4(1)(16) = 16 – 64 = -48$$ (Negative discriminant means no real solutions)

B. $$x^2 – 8x – 16 = 0$$
Discriminant: $$(-8)^2 – 4(1)(-16) = 64 + 64 = 128$$ (Positive discriminant means two distinct real solutions)

C. $$x^2 – 6x – 16 = 0$$
Discriminant: $$(-6)^2 – 4(1)(-16) = 36 + 64 = 100$$ (Positive discriminant means two distinct real solutions)

D. $$x^2 – 8x + 16 = 0$$
Discriminant: $$(-8)^2 – 4(1)(16) = 64 – 64 = 0$$ (Zero discriminant means exactly one distinct real solution)

So, the correct answer is option D, $$x^2 – 8x + 16 = 0$$.

[No calc]  Question   medium

$$3x^2 − 7x − 1 = 0$$

What are the solutions to the given equation?

A) $$x=\frac{7\pm\sqrt{37}}{6}$$

B) $$x=\frac{7\pm\sqrt{61}}{6}$$

C) $$x=\frac{-7\pm\sqrt{37}}{6}$$

D) $$x=\frac{-7\pm\sqrt{61}}{6}$$

B)  $$x=\frac{7\pm\sqrt{61}}{6}$$

To solve the quadratic equation $$3x^2 – 7x – 1 = 0$$, we use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$, where $$a = 3$$, $$b = -7$$, and $$c = -1$$.

Calculate the discriminant $$b^2 – 4ac$$:
$b^2 – 4ac = (-7)^2 – 4(3)(-1) = 49 + 12 = 61$

Substitute $$a$$, $$b$$, and the discriminant into the quadratic formula:
$x = \frac{-(-7) \pm \sqrt{61}}{2(3)} = \frac{7 \pm \sqrt{61}}{6}$

Thus, the solutions to the equation are:
$x = \frac{7 \pm \sqrt{61}}{6}$

[Calc]  Question  Medium

$$x^2 – 8x+ 20 = 0$$
How many distinct real solutions does the given equation have ?
A) Zero
B) Exactly one
C) Exactly two
D) Infinitely many

A) Zero

To determine the number of distinct real solutions to the equation $$x^2 – 8x + 20 = 0$$, we need to calculate the discriminant ($$\Delta$$).

1. Calculate the discriminant:
$\Delta = b^2 – 4ac$
For the equation $$ax^2 + bx + c = 0$$, we have:
$a = 1, \quad b = -8, \quad c = 20$
$\Delta = (-8)^2 – 4 \cdot 1 \cdot 20$
$\Delta = 64 – 80$
$\Delta = -16$

Since the discriminant is negative, there are no real solutions, only complex solutions.

[No calc]  Question  medium

The graph shown models the profit y, in thousands of dollars, and the number of products sold x, in
thousands, for a certain company. Which equation represents this model?

A) $$y=\frac{1}{18}x^2$$

B) $$y=\frac{-1}{18}x^2+50$$

C) $$y=50-\frac{1}{18}{(x-30)}^2$$

D) $$y=50+\frac{1}{18}{(x-30)}^2$$

C)  $$y=50-\frac{1}{18}{(x-30)}^2$$

Given the graph is a downward parabola with a vertex at $$(30, 50)$$ and passing through the points $$(0,0)$$ and $$(60,0)$$, we need to determine the equation of this parabola.

The general form of a parabola is:
$y = a(x – h)^2 + k$

Here, $$(h, k)$$ is the vertex of the parabola.
$h = 30$
$k = 50$

So, the equation becomes:
$y = a(x – 30)^2 + 50$

Since the parabola passes through the point $$(0, 0)$$, we can substitute $$x = 0$$ and $$y = 0$$ into the equation to find $$a$$:
$0 = a(0 – 30)^2 + 50$
$0 = a(900) + 50$
$0 = 900a + 50$
$-50 = 900a$
$a = -\frac{50}{900}$
$a = -\frac{1}{18}$

Thus, the equation of the parabola is:
$y = -\frac{1}{18}(x – 30)^2 + 50$

To verify, let’s check if the parabola passes through $$(60, 0)$$:
$y = -\frac{1}{18}(60 – 30)^2 + 50$
$y = -\frac{1}{18}(30)^2 + 50$
$y = -\frac{1}{18}(900) + 50$
$y = -50 + 50$
$y = 0$

This confirms that the point $$(60, 0)$$ lies on the parabola.

Therefore, the equation representing the model is:
$y = -\frac{1}{18}(x – 30)^2 + 50$

[Calc]  Question   medium

$${(2x+1)}^2=81$$

What are all possible solutions to the given equation?

A) −9 and 9

B) −5 and 4

C) 4

D) 40

B) −5 and 4

To find all possible solutions to the equation $$(2x + 1)^2 = 81$$:

Expand the left side of the equation:
$(2x + 1)^2 = 81$
$(2x + 1)(2x + 1) = 81$
$4x^2 + 4x + 1 = 81$

Subtract 81 from both sides:
$4x^2 + 4x + 1 – 81 = 0$
$4x^2 + 4x – 80 = 0$

Divide both sides by 4:
$x^2 + x – 20 = 0$

$(x – 4)(x + 5) = 0$

Set each factor equal to zero and solve for $$x$$:
$$x – 4 = 0$$ gives $$x = 4$$
$$x + 5 = 0$$ gives $$x = -5$$

Therefore, the possible solutions are $$x = 4$$ and $$x = -5$$.

[Calc]  Questions  Medium

$$3x^{2}+x-2$$

The solutions to the quadratic equation above are a and b. What is the value of a + b ?

A) -5

B) -1/3

C) 1/3

D) 5/3

Ans: B

To solve the quadratic equation $$3x^2+x-2=0$$, we can use the quadratic formula:
$x = \frac{{-b \pm \sqrt{{b^2 – 4ac}}}}{{2a}}$

Where $$a = 3$$, $$b = 1$$, and $$c = -2$$. Substituting these values into the quadratic formula:

$x = \frac{{-(1) \pm \sqrt{{(1)^2 – 4(3)(-2)}}}}{{2(3)}}$
$x = \frac{{-1 \pm \sqrt{{1 + 24}}}}{6}$
$x = \frac{{-1 \pm \sqrt{{25}}}}{6}$
$x = \frac{{-1 \pm 5}}{6}$

So, the solutions are $$x = \frac{{-1 + 5}}{6} = \frac{4}{6} = \frac{2}{3}$$ and $$x = \frac{{-1 – 5}}{6} = \frac{-6}{6} = -1$$.

Therefore, $$a = -1$$ and $$b = \frac{2}{3}$$.

Thus, $$a + b = -1 + \frac{2}{3} = -\frac{3}{3} + \frac{2}{3} = -\frac{1}{3}$$.

So, the correct answer is $$\mathbf{B}$$ – $$-\frac{1}{3}$$.

[No- Calc]  Question   Medium

What is the $$y$$-intercept of the graph of $$y=(x-4)^2+3$$ in the xy-plane?
A) $$(0,-13)$$
B) $$(0,0)$$
C) $$(0,3)$$
D) $$(0,19)$$

Ans:D

To find the $$y$$-intercept of the graph of $$y = (x-4)^2 + 3$$, we need to determine the point where the graph intersects the y-axis, which is when $$x = 0$$.

Substitute $$x = 0$$ into the equation:

$y = (0-4)^2 + 3$
$y = (-4)^2 + 3$
$y = 16 + 3$
$y = 19$

So, the $$y$$-intercept is $$(0, 19)$$.

D) $$(0, 19)$$

[No- Calc]  Question  Medium

$f(x)=x^2+4 x+4$

For the given function $$f$$, what is the minimum value of $$f(x)$$ ?
A) 4
B) 2
C) 1
D) 0

Ans:D

To find the minimum value of the given function $$f(x) = x^2 + 4x + 4$$, we can complete the square to express it in vertex form, $$f(x) = a(x – h)^2 + k$$, where $$(h, k)$$ is the vertex of the parabola.

Completing the square:

$f(x) = x^2 + 4x + 4$
$f(x) = (x + 2)^2$

From the vertex form, we can see that the minimum value occurs when $$x = -2$$, and at that point, $$f(x) = 0$$.

So, the minimum value of $$f(x)$$ is $$0$$.

D) 0

[Calc]  Question   Medium

$$f(x)=a(x-b)(x-c)$$

For the quadratic function $f$ shown, $a, b$, and $c$ are constants. For the graph of the $y=\mathrm{f}(x)$ in the $x y$ plane, the quadratic function $f$ opens upward, and the coordinates of its vertex are both negative. Which of the following could be true?
A) $\mathrm{a}<0, \mathrm{~b}<0, \mathrm{c}<0$
B) $\mathrm{a}<0, \mathrm{~b}>0, \mathrm{c}>0$
C) $\mathrm{a}>0, \mathrm{~b}<0, \mathrm{c}<0$
D) $\mathrm{a}>0, \mathrm{~b}>0, \mathrm{c}>0$

C

[Calc]  Question   Medium

Fahari kicks a ball on the ground into the air. One second after being kicked, the ball reaches its maximum height of 16 feet above the ground, and 2 seconds after being kicked, the ball is back on the ground. A quadratic function models the height $h(t)$, in feet, of the ball $t$ seconds after Fahari kicks it. Which equation defines this relationship?
A) $h(t)=-16(t-16)^2+1$
B) $h(t)=-16(t-1)^2+16$
C) $h(t)=-16(t+1)^2-16$
D) $h(t)=-16(t+16)^2-1$

B

[Calc]  Question Medium

$$(x-5)^2+10(x-5)+25=0$$

How many distinct real solutions does the given equation have?
A) Zero
B) Exactly one
C) Exactly two
D) Infinitely many

B

[Calc]  Question  Medium

$$x^2-14 x+40=2 x+1$$

What is the sum of the solutions to the given equation?
A) -16
B) -14
C) 14
D) 16

D

[Calc]  Question Medium

$$5(x-3)(x+1)=0$$

What positive value of $x$ satisfies the equation above?

3

[Calc]  Question Medium

$$x^2+b x+16=0$$

In the quadratic equation shown, $b$ is a constant. For what values of $b$ does the equation have only one solution?
A) -4 only
B) -8 only
C) -4 and 4
D) -8 and 8

D

[Calc]  Question Medium

$$h(x)=(x-5)(x+5)$$

The function $h$ is defined as shown. For what value of $x$ does the function $h$ reach its minimum value?
A) -25
B) -5
C) 0
D) 5

C

Question

The graph of the function $$f$$ defined by $$f(x)=\frac{1}{4}(x+4)^{2}-5$$ is shown. The graph passes through the points (-10, $$a$$) and ($$a$$, $$b$$), where $$a$$ and $$b$$ are constants. What is the value of $$b$$ ?

1. 4
2. 9
3. 11
4. 16

C

Question

$$x$$2-12$$x$$+23=0

Which of the following is a solution to the given equation?

1. -12+$$\sqrt{11}$$
2. 12+$$\sqrt{11}$$
3. -6+$$\sqrt{13}$$
4. 6+$$\sqrt{13}$$

D

Question

A potter is selling vases. The function $$p$$ gives the total profit $$p(x)$$, in dollars, the potter will receive if the vases are sold at a price of $$x$$ dollars each. The graph of $$y = p(x)$$ is shown.

What equation represents the relationship between the price of one vase and the profit?

1. $$p(x)=-6(x-18)^{2} +34$$
2. $$p(x) = -6(x – 18)^{2} + 1536$$
3. $$p(x) = -6(x – 34)^{2}+ 18$$
4. $$p(x) = -6(x – 34)^{2}+ 1536$$

B

Question

In the given quadratic function , is a constant and . What is the value of ?

1. 8
2. 30
3. 32
4. 468

A

Question

The graph above models the daily profit $$y$$, in dollars, that a school club expects to make from selling shirts for a price of $$x$$ dollars. Based on the model, what is the maximum daily profit, in dollars? (Disregard the  sign when gridding your answer.)

Ans: 60

Questions

$x-y=1$
$x+y=x^2-3$

Which ordered pair is a solution to the system of equations above?
A. $(1+\sqrt{3}, 3)$
B. $(\sqrt{3},-\sqrt{3})$
C. $(1+\sqrt{5}, \sqrt{5})$
D. $(\sqrt{5},-1+\sqrt{5})$

Ans: A

Questions

$4 v^2+6 v+1=0$

Which of the following values is a solution to the equation above?
A. $\frac{-3+\sqrt{5}}{4}$
B. $\frac{-3+\sqrt{13}}{4}$
C. $\frac{3+\sqrt{5}}{4}$
D. $\frac{3+\sqrt{13}}{4}$

Ans: A

Questions

If $7(2 x-5)-2(2 x-5)=4(x+5)$, what is the value of $x$ ?
A. 1
B. $\frac{15}{2}$
C. $\frac{65}{6}$
D. 65

Ans: B

Question

For the function $$f$$, the table above shows some values of $$x$$ and their corresponding values of $$f(x)$$ in terms of the constants $$a, b$$, and $$c$$. If $$a < b < c$$, which of the following could NOT be the graph of $$y = f (x)$$ in the $$xy$$-plane?

Ans: B

Question

$f(n)=5.77\left(0.98^n\right)$

The function above can be used to estimate the number of farms, $f(n)$, in millions, in the United States for $0 \leq n \leq 72$, where $n$ is the number of years after 1940. Which of the following is the best interpretation of the number 5.77 in this context?
A. The estimated number of farms, in millions, in 1940
B. The estimated number of farms, in millions, $n$ years after 1940
C. The estimated decrease in the number of farms, in millions, each year after 1940
D. The estimated percent by which the number of farms decreased from each year to the next after 1940

Ans: A

Questions

The gas mileage $M(s)$, in miles per gallon, of a car traveling $s$ miles per hour is modeled by the function below, where $20 \leq s \leq 75$. $M(s)=-\frac{1}{24} s^2+4 s-50$
According to the model, at what speed, in miles per hour, does the car obtain its greatest gas mileage?
A. 46
B. 48
C. 50
D. 75

Ans: B

Question

$y=x^2 y=2 x+3$ The system of equations above is graphed in the $x y$-plane. The graphs of the equations intersect at a point $(x, y)$ where $x>0$ and $y>0$. What is the $y$-coordinate of this point of intersection?
A. 1
B. 3
C. 5
D. 9

Ans: D

Question

$k x+y=1 y=-x^2+k$

In the system of equations above, $k$ is a constant. When the equations are graphed in the $x y$-plane, the graphs intersect at exactly two points. Which of the following CANNOT be the value of $k$ ?
A. 3
B. 2
C. 1
D. 0

Ans: D

Questions

$$y =-x^2 +4x$$
Which of the following is the graph in the $$xy$$-plane of the given equation?

Ans: D

Questions

$\frac{1}{x^2+10 x+25}=4$

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. $\frac{11}{2}$

Ans: A

Questions

$2 x^2-2=2 x+3$ Which of the following is a solution to the equation above?
A. 2
B. $1-\sqrt{11}$
C. $\frac{1}{2}+\sqrt{11}$
D. $\frac{1+\sqrt{11}}{2}$

Ans: D

Questions

$P(x)=x^2-11 x+k$

In the function above, $k$ is a constant. If 2 is a zero of the function, what is the value of $k ?$
A. -18
B. -2
C. 3
D. 18

Ans: D

Questions

$y=x^2-6 x-16$

The graph of the equation above in the $x y$-plane is a parabola. Which of the following equivalent forms of the equation includes the $x$ – and $y$ coordinates of the vertex as constants?
A. $y=(x-3)^2-25$
B. $y=x(x-6)-16$
C. $y=x^2-2(3 x+8)$
D. $y+16=x(x-6)$

Ans: A

Questions

Which of the following is a solution to the equation $\sqrt{14-x}+2=x$
I. -2
II. 1
III. 5
A. I only
B. II only
C. III only
D. I and III