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Applications of Quadratics: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A number and its square

Three times the square of a positive number is $20$ more than seven times that number. Write an equation in standard form for the number $n$ ; do not solve it.
1. Hint 1
  Each described quantity has to be written with the one letter that names the unknown.
2. Hint 2
  Turn the comparison into an equation, then gather every term on one side so the other side is zero.
Answer

$3n^2-7n-20=0$ , or equivalently $-3n^2+7n+20=0$ .

Full solution

Three times the square of the number is $3n^2$ , and seven times the number is $7n$ .
The comparison says
$3n^2=7n+20$

Move both right-hand terms across, leaving zero on the right.
$3n^2-7n-20=0$

The number is described as positive, so the equation is read with $n>0$ .

Answer

$3n^2-7n-20=0$ , or equivalently $-3n^2+7n+20=0$ .

Key idea

Translating a comparison means writing every quantity with one unknown and gathering the terms on one side.
Problem 1: Mark complete
Problem 2 A height record

A ball is launched upward from $12$ feet above the ground. After $1$ second it is $20$ feet above the ground. In the model $h=-16t^2+v_0t+h_0$ , find the initial upward speed $v_0$ in feet per second.
1. Hint 1
  The starting height supplies the constant in the model.
2. Hint 2
  Substitute the recorded time and height, then isolate the speed.
Answer

$24$ feet per second.

Full solution

The record gives
$20=-16(1)^2+v_0(1)+12$

Collect the constants and solve.
$v_0=24$

Check: after one second the model gives $-16+24+12=20$ feet.

Answer

$24$ feet per second.

Key idea

A recorded height and time can determine a missing coefficient in a projectile model.
Problem 2: Mark complete
Problem 3 Two areas

A square has side length $s$ centimeters. A triangle has base $6$ centimeters and height $s$ centimeters. The square has four times the triangle area. Find the side length $s$ .
1. Hint 1
  Write each area using the same unknown length.
2. Hint 2
  Set the two areas in the stated ratio and solve the equation that results.
Answer

$s=12$ centimeters.

Full solution

The triangle area is $3s$ square centimeters.
The comparison gives
$s^2=12s$

Bring both terms to one side and factor out the common side length.
$s(s-12)=0$

The roots are $0$ and $12$ .
A side length must be positive, so keep $12$ .
The square area is $144$ and the triangle area is $36$ square centimeters, confirming the ratio.

Answer

$s=12$ centimeters.

Key idea

A comparison of areas can yield a zero root that the geometric situation excludes.
Problem 3: Mark complete
Problem 4 A rising platform

A ball starts $24$ feet above the ground with initial upward speed $40$ feet per second. At the same instant, a platform starts at the same height and rises steadily at $8$ feet per second. Using the projectile model $h=-16t^2+v_0t+h_0$ for the ball, when after the start are the ball and platform at the same height again?
1. Hint 1
  Write a height expression for each moving object.
2. Hint 2
  Equate the heights and distinguish the starting instant from a later meeting.
Answer

$2$ seconds after the start.

Full solution

The ball height is $-16t^2+40t+24$ and the platform height is $8t+24$ , in feet.
Equality gives
$-16t^2+32t=0$

Factor the common time.
$-16t(t-2)=0$

The roots are $0$ and $2$ .
The question asks for a later meeting, so the time is $2$ seconds.
At that instant both heights are $40$ feet.

Answer

$2$ seconds after the start.

Key idea

Matching two changing heights means equating their expressions and interpreting each resulting time.
Problem 4: Mark complete
Problem 5 A trimmed square

A square sheet measures $10$ centimeters on each side. A strip of width $x$ is removed along one edge, and a strip of width $2x$ is removed along an adjacent edge. The remaining rectangle has area $48$ square centimeters. Find every value of $x$ the sheet allows, and justify any root you discard.
1. Hint 1
  The two remaining side lengths lose different amounts.
2. Hint 2
  Multiply those lengths, solve the quadratic, and test each root against the sheet's $10$ centimeter sides.
Answer

$x=2$ centimeters; the root $x=13$ is rejected: it would leave side lengths of $-3$ and $-16$ centimeters.

Full solution

The remaining dimensions are $10-x$ and $10-2x$ centimeters.
$(10-x)(10-2x)=48$

Expand, move $48$ across, and divide by $2$ .
$x^2-15x+26=0$

Factor.
$(x-2)(x-13)=0$

The root $13$ would leave side lengths of $-3$ and $-16$ centimeters, which no sheet can have, so it is rejected.
The root $2$ leaves side lengths $8$ and $6$ centimeters, whose product is $48$ square centimeters.

Answer

$x=2$ centimeters; the root $x=13$ is rejected: it would leave side lengths of $-3$ and $-16$ centimeters.

Key idea

Every side remaining after a cut must stay positive when an algebraic root is interpreted.
Problem 5: Mark complete
Problem 6 A triangular support

A right triangle has legs $x$ centimeters and $2x$ centimeters. Its hypotenuse is $x+6$ centimeters. Find the positive value of $x$ exactly.
1. Hint 1
  The hypotenuse square equals the sum of the leg squares.
2. Hint 2
  After expanding, solve the quadratic and reject a nonpositive side length.
Answer

$x=\dfrac{3+3\sqrt5}{2}$ centimeters.

Full solution

The right triangle relation is
$x^2+(2x)^2=(x+6)^2$

Expanding and dividing by $4$ gives
$x^2-3x-9=0$

The quadratic formula gives
$x=\frac{3\pm\sqrt{45}}2$

The negative choice is below zero.
The positive choice is $(3+3\sqrt5)/2$ .
For this value, $x^2=3x+9$ , so $5x^2=(x+6)^2$ , checking the side relation.

Answer

$x=\dfrac{3+3\sqrt5}{2}$ centimeters.

Key idea

A right triangle model must satisfy both the squared side relation and positive side lengths.
Problem 6: Mark complete
Problem 7 A sales target

A vendor charges $p$ dollars per item and sells $60-3p$ items, where $p$ is a whole number with $5<p<20$ . Each item costs the vendor $5$ dollars, and the total profit is the number of items sold times the amount by which the price exceeds that cost. Find every permitted price that gives a total profit of $108$ dollars.
1. Hint 1
  Both the number sold and the profit on one item change with the price, so the condition is a quadratic in $p$ .
2. Hint 2
  Multiply profit per item by the number sold, then test both roots against the price restriction.
Answer

$8$ dollars or $17$ dollars per item.

Full solution

The total profit condition is
$(p-5)(60-3p)=108$

Expand, gather terms, and divide by $-3$ .
$p^2-25p+136=0$

Factor.
$(p-8)(p-17)=0$

Both prices are permitted whole numbers.
At $8$ dollars, profit is $3(36)=108$ dollars; at $17$ dollars, it is $12(9)=108$ dollars.

Answer

$8$ dollars or $17$ dollars per item.

Key idea

Total profit multiplies the profit on one item by the number sold, and both algebraic prices may be permitted.
Problem 7: Mark complete
Problem 8 Two launch sites

Two balls are launched with the same initial upward speed of $56$ feet per second. One starts at ground level and the other starts $26$ feet above the ground. A student says that if the lower ball never reaches $50$ feet, neither does the higher ball. Decide whether the claim is correct by examining the equations for height $50$ in the model $h=-16t^2+v_0t+h_0$ .
1. Hint 1
  The two models differ in their starting-height constants.
2. Hint 2
  Compare the discriminants after setting each height to the target, and check that any real times are positive.
Answer

False; only the ball launched from $26$ feet reaches $50$ feet, at $0.5$ second and at $3$ seconds.

Full solution

Set the lower ball's height to the target.
$-16t^2+56t=50$

Move $50$ across and divide by $-2$ .
$8t^2-28t+25=0$

Its discriminant is
$D=784-800$

This is $-16$ , so no real time reaches the target.
Now set the higher ball's height to the target.
$-16t^2+56t+26=50$

Move $50$ across and divide by $-8$ .
$2t^2-7t+3=0$

Its discriminant is $49-24=25$ , which is positive, and the quadratic factors as
$(2t-1)(t-3)=0.$
The times are $0.5$ second and $3$ seconds, both positive and before the ball lands.
Therefore the higher starting point changes whether the target is reached.

Answer

False; only the ball launched from $26$ feet reaches $50$ feet, at $0.5$ second and at $3$ seconds.

Key idea

Changing a starting height changes the discriminant of a target-height equation even when launch speeds match.
Problem 8: Mark complete
Problem 9 A tile order

A square arrangement uses the same whole number $n$ of tiles in each row and column, with $n>0$ . An order contains exactly the tiles for the square, one extra row of $n$ tiles, and $6$ spare tiles, for $30$ tiles altogether. A student concludes that no such arrangement is possible. Is the conclusion correct? Explain.
1. Hint 1
  Write the total count as the square count plus the extra row plus the spares.
2. Hint 2
  Bring every term to one side and solve the quadratic, then read each root against what $n$ stands for.
Answer

Yes; no positive whole number $n$ satisfies the order.

Full solution

The total count gives
$n^2+n+6=30$

Move $30$ across.
$n^2+n-24=0$

Its discriminant is $1+96=97$ , so the quadratic formula gives
$n=\frac{-1\pm\sqrt{97}}{2}$

Since $81<97<100$ , the positive root lies between $4$ and $4.5$ , so it is not a whole number, and the negative root is excluded by $n>0$ .
Checking directly, $n=4$ needs $26$ tiles and $n=5$ needs $36$ tiles, so no square arrangement fits an order of $30$ .

Answer

Yes; no positive whole number $n$ satisfies the order.

Key idea

A positive real root need not be a valid count when the unknown must be a whole number.
Problem 9: Mark complete
Problem 10 A change in size

A square originally has side length $8$ centimeters. Each side changes by $x$ centimeters, where a negative $x$ means a decrease. The new area is $25$ square centimeters. Solving the area equation gives $x=-3$ and $x=-13$ . A student rejects both because they are negative. Is that correct? Explain which changes are permitted.
1. Hint 1
  The unknown describes a change, not the final side length.
2. Hint 2
  Check the resulting side length for each candidate change.
Answer

No; $x=-3$ centimeters is permitted, and $x=-13$ centimeters is not.

Full solution

The final side length is $8+x$ , so the equation is
$(8+x)^2=25$

So $8+x=5$ or $8+x=-5$ , giving $x=-3$ or $x=-13$ , the two changes the statement reports.
For $x=-3$ , the actual side is $5$ centimeters, which has area $25$ square centimeters.
For $x=-13$ , the proposed side is $-5$ centimeters, which is not a length.
The negative change $-3$ is therefore valid.

Answer

No; $x=-3$ centimeters is permitted, and $x=-13$ centimeters is not.

Key idea

A negative change may be meaningful when it leaves the actual measured quantity positive.
Problem 10: Mark complete

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