VA SOL Algebra I Practice Questions (2026) — Free Sample Questions with Answers

Answer: A

1. 1. $\sqrt{50}=5\sqrt{2}$, $\sqrt{18}=3\sqrt{2}$, $\sqrt{8}=2\sqrt{2}$.
2. 2. Combine like radicals: $(5-3+2)\sqrt{2}=4\sqrt{2}$.
3. 3. Check: $7.071-4.243+2.828=5.657$ and $4\sqrt{2}\approx 5.657$.

$4\sqrt{2}$: Correct. $5\sqrt{2}-3\sqrt{2}+2\sqrt{2}=4\sqrt{2}$ (A.EO.4.c).

$2\sqrt{10}$: Combines the radicands, $\sqrt{50-18+8}=\sqrt{40}$; square roots cannot be added or subtracted that way.

$6\sqrt{2}$: Writes $\sqrt{8}$ as $4\sqrt{2}$, taking out 4 instead of $\sqrt{4}=2$.

$10\sqrt{2}$: Adds $3\sqrt{2}$ instead of subtracting it.

Rewrite each square root in simplest form to reveal like radicals, then combine only the coefficients.

Answer: A

1. 1. Outer area: $(2x+5{)}^2=4x^2+10x+10x+25=4x^2+20x+25$.
2. 2. Patio area: $(2x{)}^2=4x^2$.
3. 3. Walkway: $4x^2+20x+25-4x^2=20x+25$.
4. 4. Check with $x=5$: outer $15^2=225$, patio $10^2=100$, difference 125; $20(5)+25=125$.

$20x+25$: Correct. $(2x+5{)}^2-(2x{)}^2=4x^2+20x+25-4x^2=20x+25$ (A.EO.2.a, A.EO.2.b).

$25$: Squares $2x+5$ as $4x^2+25$ and drops the middle term $2(2x)(5)=20x$.

$10x+25$: Uses only one product $(2x)(5)$ for the middle term; $(a+b{)}^2$ has the middle term $2ab$.

$8x^2+20x+25$: Adds the patio area to the outer square instead of subtracting it.

The area of a border is the outer area minus the inner area; when squaring a binomial, keep the middle term $2ab$.

Answer: C

$(4x+6)(3x-5)$: Equivalent, but not complete: the factor $4x+6$ still contains the common factor 2.

$2(2x-3)(3x+5)$: Signs are switched: this expands to $2(6x^2+x-15)$, giving $+2x$ instead of $-2x$.

$2(2x+3)(3x-5)$: Correct. The GCF is 2: $12x^2-2x-30=2(6x^2-x-15)$, and $6x^2-x-15=(2x+3)(3x-5)$ because $-10x+9x=-x$ (A.EO.2.c).

$2(6x+5)(x-3)$: Wrong factor pair: $(6x+5)(x-3)=6x^2-13x-15$, so the middle term is wrong.

Factor out the greatest common factor first, then factor the remaining trinomial; a factorization is complete only when no factor can be factored further.

Answer: D

1. Shop P: $y=3x+20$; Shop Q: $y=5x+8$.
2. Set equal: $3x+20=5x+8\Rightarrow 12=2x\Rightarrow x=6$.
3. $y=3(6)+20=38$, so the lines meet at $(6,38)$.
4. Check: $5(6)+8=38$ ✓.

For a 38-hour rental, both shops charge a total of $6.: Reverses the coordinates: $x$ is hours and $y$ is cost, so the point $(6,38)$ means 6 hours and $38.

Both shops charge the same hourly rate of $6 per hour.: Treats the x-coordinate as a rate; the hourly rates are $3 and $5 and never become equal.

Shop P costs less than Shop Q for any rental of 6 hours or less.: Reverses the comparison: below 6 hours Shop Q's line is lower, so Shop Q costs less.

A 6-hour rental costs the same at both shops, $38 in total.: Correct. The lines meet at $(6,38)$: after 6 hours both totals are $38, since $20+3(6)=8+5(6)=38$.

The intersection of two models is the input where their outputs are equal; state both coordinates with their units (A.EI.2.h).

Answer: C

1. $6x-4y=10\Rightarrow -4y=-6x+10\Rightarrow y=1.5x-2.5$.
2. Compare with $y=1.5x+2.5$: same slope $1.5$, different y-intercepts.
3. Parallel, distinct lines have no point in common.
4. Check algebraically: substituting gives $6x-4(1.5x+2.5)=-10$, and $-10=10$ is false for every $x$.

It has exactly one solution, because the slopes of the lines are $-1.5$ and $1.5$.: Takes the slope of $6x-4y=10$ as $\frac{A}{B}=\frac{6}{-4}$; solving for $y$ gives a slope of $1.5$, the same as the other line.

It has exactly one solution, because the y-intercepts of the lines are different.: Different y-intercepts alone do not make lines cross; lines with equal slopes and different intercepts never meet.

It has no solution, because the lines have the same slope and different y-intercepts.: Correct. $6x-4y=10$ becomes $y=1.5x-2.5$, so both slopes are $1.5$ and the intercepts are $-2.5$ and $2.5$: parallel lines that never intersect.

It has infinitely many solutions, because both equations describe the same line.: Results from a sign slip when solving for $y$ ($y=1.5x+2.5$); dividing $-6x+10$ by $-4$ gives $1.5x-2.5$.

Rewrite both equations in slope-intercept form: different slopes give one solution; equal slopes with different intercepts give none; identical lines give infinitely many (A.EI.2.c).

Answer: C

1. 1. Multiply both sides by 6: $4(x-6)=3x+6$.
2. 2. Distribute: $4x-24=3x+6$.
3. 3. $x=30$.
4. 4. Check: $\frac{2}{3}(24)=16$ and $\frac{1}{2}(30)+1=16$.

$-18$: Reaches $4x-24=3x+6$ but subtracts 24 instead of adding it when isolating x.

$25$: Multiplies the fraction terms by 6 but leaves the constant 1 unchanged: $4x-24=3x+1$.

$30$: Correct. Multiply by 6: $4(x-6)=3x+6$, so $4x-24=3x+6$ and $x=30$ (A.EI.1.b).

$42$: Multiplies only x by $\frac{2}{3}$, writing $\frac{2}{3}x-6$, which leads to $4x-36=3x+6$.

Clear fractions by multiplying every term by the LCD, and distribute the coefficient to both terms inside the parentheses.

Answer: B

1. 1. $ky-5y=m+3$.
2. 2. Factor out y: $y(k-5)=m+3$.
3. 3. $y=\frac{m+3}{k-5}$.
4. 4. Check with $k=7$, $m=1$: $7y-3=5y+1\Rightarrow y=2$, and $\frac{1+3}{7-5}=2$.

$y=\frac{m+3}{k+5}$: Moves $5y$ to the left without changing its sign; subtracting $5y$ gives $k-5$.

$y=\frac{m+3}{k-5}$: Correct. Collect the y-terms: $ky-5y=m+3$; factor: $y(k-5)=m+3$; divide by $k-5$ (A.EI.1.d).

$y=\frac{m-3}{k-5}$: Moves $-3$ to the right without changing its sign; adding 3 gives $m+3$.

$y=\frac{m+3}{k}-5$: Divides by k before collecting the y-terms; y appears in two terms, so they must be factored first.

When the target variable appears in more than one term, collect those terms on one side and factor out the variable before dividing.

Answer: D

1. 1. Set $h=0$: $-16t^2+24t+40=0$.
2. 2. Divide by $-8$: $2t^2-3t-5=0$.
3. 3. Factor: $(2t-5)(t+1)=0$, so $t=2.5$ or $t=-1$; reject $-1$.
4. 4. Check: $-16(6.25)+24(2.5)+40=-100+60+40=0$.

0.75: This is the time of the maximum height, $t=-\frac{24}{2(-16)}$, not the time when $h=0$.

1: Factors with the signs reversed, $(2t+5)(t-1)$, which expands to $2t^2+3t-5$.

1.5: Solves $-16t^2+24t=0$, the time the ball returns to its starting height of 40 feet.

2.5: Correct. $-8(2t^2-3t-5)=-8(2t-5)(t+1)=0$, so $t=2.5$ or $t=-1$; time cannot be negative (A.EI.3.a).

"Reaches the ground" means the output is 0; solve the quadratic and reject the negative time, and do not confuse the zero with the vertex.

Answer: B

1. 1. Standard form: $4x^2-12x+9=0$, so $a=4$, $b=-12$, $c=9$.
2. 2. $b^2-4ac=(-12{)}^2-4(4)(9)=144-144=0$.
3. 3. Discriminant 0 means exactly one real solution: $(2x-3{)}^2=0$, $x=\frac{3}{2}$.
4. 4. Check: $4\left(\frac{9}{4}\right)+9=18$ and $12\left(\frac{3}{2}\right)=18$.

No real solutions, because the discriminant is $-144$: Uses $b=0$ by not moving $12x$ to the left side first; in standard form $b=-12$.

One real solution, because the discriminant is 0: Correct. In standard form, $4x^2-12x+9=0$, so $b^2-4ac=144-144=0$; the solution is $x=\frac{3}{2}$ (A.EI.3.b).

Two real solutions, because the discriminant is 288: Computes $b^2+4ac$; the discriminant is $b^2-4ac$.

Two real solutions, because the discriminant is 108: Leaves out the 4 in $4ac$: $144-36=108$.

Write the equation in standard form before reading a, b and c; a positive discriminant gives two real solutions, zero gives one, and a negative one gives none.

Answer: D

1. Substitute $t=0$: $d=1.05(0{)}^2+1.49(0)+3.89=3.89$.
2. So 3.89 is the predicted distance, in centimeters, when $t=0$, the start of timing.
3. Check against the table: at 0 seconds Vivian measured 4 cm, close to 3.89.

The distance the ball travels during each second: Treats the constant term as a rate; a constant amount per second would be a slope in a linear model, and here the distance gained each second keeps growing.

The time, in seconds, when the ball reaches its greatest distance: Confuses the constant term with a vertex and mixes up the variables; 3.89 is a distance, and this curve keeps increasing over the data.

The total distance the ball travels in the first 2 seconds: Reads 3.89 as an interval total; the model predicts about 11.07 cm at 2 seconds, so the ball travels about 7 cm in that time.

The predicted distance from the mark at 0 seconds: Correct. At $t=0$ the model gives $d=3.89$, so 3.89 is the predicted distance from the starting mark when timing begins, close to the measured 4 cm (A.ST.1.e).

The constant term of a curve of best fit is the predicted output when the input is 0; interpret it in the units of the context (A.ST.1.e).

Answer: A

1. Table: $x$: 0, 1, 2, 3, 4, 5, 6.
2. $g(x)=x^2$: 0, 1, 4, 9, 16, 25, 36.
3. $h(x)=2^x$: 1, 2, 4, 8, 16, 32, 64.
4. $g>h$ only at $x=3$; they tie at $x=2$ and $x=4$; $h>g$ at $x=0,1,5,6$.

$h(x)$ is greater than $g(x)$ at $x=6$ but not at $x=3$.: Correct. $g(3)=9>h(3)=8$, but $h(6)=64>g(6)=36$; the exponential overtakes the quadratic.

$g(x)$ is greater than $h(x)$ for every value of $x$ from 0 to 6.: Fails at both ends: $h(0)=1>g(0)=0$, and $h(6)=64>g(6)=36$.

$h(x)$ is greater than $g(x)$ for every value of $x$ from 0 to 6.: Fails at $x=3$, where $g(3)=9$ and $h(3)=8$, and at $x=4$, where both equal 16.

$f(x)$ is greater than $g(x)$ for every value of $x$ from 1 to 6.: $f(1)=g(1)=1$, and for $x>1$, $x^2>x$, so $g$ is greater, not $f$.

An increasing exponential function eventually exceeds any quadratic or linear function, even if it is smaller for some inputs (A.F.2.h).

Answer: D

1. Set $h(t)=0$: $-16t^2+64t+80=0\Rightarrow -16(t^2-4t-5)=0$.
2. Factor: $-16(t-5)(t+1)=0$, so $t=5$ or $t=-1$.
3. Reject $t=-1$ (before launch). The rocket is in the air from $t=0$ to $t=5$.
4. Check: $h(5)=-400+320+80=0$ ✓.

$t\ge 0$: Has no upper limit; after the rocket lands at $t=5$, the model no longer describes its height.

$-1\le t\le 5$: Keeps the negative zero, $t=-1$; time before the launch is not part of this situation.

$0\le t\le 2$: Stops at the vertex, $t=2$, when the rocket reaches its highest point; it is still in the air until $t=5$.

$0\le t\le 5$: Correct. Time starts at launch, $t=0$, and ends when the height returns to 0, at $t=5$.

In context, the domain runs from the starting input to the input where the model stops applying; reject zeros that make no sense, such as negative time (A.F.2.b).

Answer: B

1. I: slope $-0.8$, y-intercept 8.
2. II: $x=0\Rightarrow 5y=40\Rightarrow y=8$; $y=0\Rightarrow 4x=40\Rightarrow x=10$.
3. III: slope $-0.8$, point $(5,4)$; check $-0.8(5)+8=4$ ✓, so all three describe the same line.

Only equation I shows the slope of the line without any rewriting.: Equation III also shows the slope: in point-slope form the coefficient $-0.8$ is the slope.

I and III both show the slope; II gives both intercepts directly.: Correct. $-0.8$ appears in I and III, and setting $x=0$ or $y=0$ in II gives the intercepts $(0,8)$ and $(10,0)$ in one step.

Equation III shows that the y-intercept of the line is $(0,4)$.: Reads the point $(5,4)$ in point-slope form as the y-intercept; the y-intercept is $(0,8)$.

Equation II shows that the slope of the line is $\frac{4}{5}$.: Takes the slope from $Ax+By=C$ as $\frac{A}{B}$; the slope is $-\frac{A}{B}=-\frac{4}{5}=-0.8$.

Slope-intercept form shows m and b, point-slope form shows m and one point, and standard form gives both intercepts directly by setting x = 0 or y = 0 (A.F.1.c).

Answer: A

1. $2x-3y=9\Rightarrow -3y=-2x+9\Rightarrow y=\frac{2}{3}x-3$; slope $\frac{2}{3}$.
2. Use $(6,-1)$: $-1=\frac{2}{3}(6)+b=4+b$, so $b=-5$.
3. Equation: $y=\frac{2}{3}x-5$.
4. Check: $\frac{2}{3}(6)-5=-1$ ✓, and the slope matches the given line, with a different y-intercept.

$y=\frac{2}{3}x-5$: Correct. $2x-3y=9$ has slope $\frac{2}{3}$; through $(6,-1)$: $-1=\frac{2}{3}(6)+b$, so $b=-5$.

$y=-\frac{3}{2}x+8$: Uses the perpendicular slope, the opposite reciprocal; parallel lines have equal slopes.

$y=-\frac{2}{3}x+3$: Reads the slope as $-\frac{A}{B}$ with the wrong sign: $-\frac{2}{-3}=+\frac{2}{3}$.

$y=\frac{2}{3}x-3$: This is the given line itself; $\frac{2}{3}(6)-3=1$, not $-1$, so it does not pass through $(6,-1)$.

Parallel lines share a slope: find it by solving the given equation for y, then use the new point to find b (A.F.1.e).