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Introduction to 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 tidy rearrangement

Write $5x^2=8-3x$ in standard form, then state $a$ , $b$ and $c$ .
1. Hint 1
  Standard form keeps every term on one side, with the other side equal to zero.
2. Hint 2
  After moving the terms across, write them in order of decreasing power before reading off the three coefficients.
Answer

$5x^2+3x-8=0$ , with $a=5$ , $b=3$ and $c=-8$ ; the equivalent $-5x^2-3x+8=0$ gives $a=-5$ , $b=-3$ and $c=8$ .

Full solution

Add $3x$ to both sides and subtract $8$ from both sides, so that the right side becomes zero.
$5x^2+3x-8=0$
The terms already stand in order of decreasing power, the squared term first.
Reading the coefficients off gives $a=5$ , $b=3$ and $c=-8$ .
With $a=5$ the $x^2$ term is still there after the rearrangement, so this is not a linear equation in disguise.
Collecting on the other side instead gives $-5x^2-3x+8=0$ , the same equation with every sign reversed, so $a=-5$ , $b=-3$ and $c=8$ is equally correct.

Answer

$5x^2+3x-8=0$ , with $a=5$ , $b=3$ and $c=-8$ ; the equivalent $-5x^2-3x+8=0$ gives $a=-5$ , $b=-3$ and $c=8$ .

Key idea

Standard form collects every term on one side so that the three coefficients can be read off directly.
Problem 1: Mark complete
Problem 2 A product of two factors

Solve $(4x-3)(2x+5)=0$ .
1. Hint 1
  A product of two numbers is zero only when one of the numbers is itself zero.
2. Hint 2
  Set each factor equal to zero in turn, then solve the two linear equations that result.
Answer

$x=\tfrac34$ (or $0.75$ ) or $x=-\tfrac52$ (or $-2.5$ ).

Full solution

The left side is a product of two factors, so at least one of them is zero.
Setting the first factor to zero gives
$4x-3=0.$
Adding $3$ to both sides gives $4x=3$ , so
$x=\tfrac34.$
Setting the second factor to zero gives
$2x+5=0.$
Subtracting $5$ from both sides gives $2x=-5$ , so
$x=-\tfrac52.$
Substituting $\tfrac34$ makes the first factor zero and substituting $-\tfrac52$ makes the second factor zero, so each value turns the whole product into zero.

Answer

$x=\tfrac34$ (or $0.75$ ) or $x=-\tfrac52$ (or $-2.5$ ).

Key idea

When a product is zero, at least one factor is zero, so set each factor to zero in turn and solve it as its own linear equation.
Problem 2: Mark complete
Problem 3 Three candidates

Decide by substitution which of $x=-4$ , $x=\tfrac32$ and $x=3$ are roots of $2x^2+5x-12=0$ .
1. Hint 1
  A root is a value that makes the left side work out to exactly zero.
2. Hint 2
  Put each candidate in place of $x$ , squaring before multiplying, and compare the result with zero.
Answer

$x=-4$ and $x=\tfrac32$ are roots; $x=3$ is not.

Full solution

Substitute $x=-4$ , remembering that squaring a negative gives a positive.
$2(-4)^2+5(-4)-12=0$
The terms here are $32$ , $-20$ and $-12$ , which add to zero, so $-4$ is a root.
Now substitute $x=\tfrac32$ , for which $2(\tfrac32)^2=\tfrac92$ .
$\tfrac92+\tfrac{15}2-12=0$
So $\tfrac32$ is a root too.
Finally substitute $x=3$ .
$2(3)^2+5(3)-12=21$
That is not zero, so $3$ is not a root.

Answer

$x=-4$ and $x=\tfrac32$ are roots; $x=3$ is not.

Key idea

Substitution tests a candidate but does not find roots, so test every value you are offered.
Problem 3: Mark complete
Problem 4 No middle term

Solve $3x^2-54=0$ .
1. Hint 1
  The equation carries no $x$ term, so the squared term can be left standing alone and the square undone directly.
2. Hint 2
  Move the constant across, then divide both sides by $3$ so that $x^2$ stands by itself.
3. Hint 3
  Take square roots of both sides, keep both signs, and simplify the radical by pulling out its largest square factor.
Answer

$x=3\sqrt2$ or $x=-3\sqrt2$ , that is $x=\pm3\sqrt2$ ; equivalently $x=\pm\sqrt{18}$ .

Full solution

Add $54$ to both sides, which leaves the squared term on its own side.
$3x^2=54$
Divide both sides by $3$ .
$x^2=18$
Take square roots of both sides, keeping both signs, and simplify $\sqrt{18}$ by writing $18$ as $9\cdot2$ .
$x=\pm3\sqrt2$
To check, squaring $3\sqrt2$ gives $9\cdot2=18$ , and $-3\sqrt2$ has that same square, so either value gives
$3(18)-54=0.$

Answer

$x=3\sqrt2$ or $x=-3\sqrt2$ , that is $x=\pm3\sqrt2$ ; equivalently $x=\pm\sqrt{18}$ .

Key idea

With no $x$ term present, a quadratic comes down to a square-root step, and both signs are kept.
Problem 4: Mark complete
Problem 5 An equation setting

Find the value of $m$ that makes $(m+1)x^2+4x=6x^2-9$ linear once it is fully simplified.
1. Hint 1
  Bring every term to one side, then compare the two squared terms.
2. Hint 2
  The coefficient of $x^2$ has to become zero, while the coefficient of $x$ must stay nonzero.
Answer

$m=5$ .

Full solution

Subtract $6x^2$ from both sides and add $9$ to both sides, then collect the squared terms.
$(m-5)x^2+4x+9=0$
The squared term vanishes exactly when $m-5=0$ , so $m=5$ .
What is left is $4x+9=0$ , a linear equation with the single solution
$x=-\tfrac94.$
Any other value of $m$ leaves a nonzero squared term, so the equation stays quadratic.

Answer

$m=5$ .

Key idea

An equation is quadratic only when its squared-term coefficient is still nonzero after full simplification.
Problem 5: Mark complete
Problem 6 A graph collection

The figure shows parts of three parabolas, with arrows indicating that the curves continue. For each graph, state how many real roots the quadratic equation it comes from has, that is, how many values of $x$ give $y=0$ .

Parts of three parabolas, labeled A, B and C.

Text description of this figure

Three separate coordinate panels stand side by side, labeled A, B and C. In every panel the horizontal x-axis runs from negative five to five and the vertical y-axis runs from negative four to five, with tick marks at every whole number, equal unit lengths on both axes, and the origin labeled 0. Panel A shows part of a curve that opens upward; its lowest point is one unit left of the vertical axis and one unit above the horizontal axis, and its two branches rise to the top edge of the panel, each ending in an arrow. Panel B shows part of a curve that opens downward; its highest point is one unit right of the vertical axis and four units above the horizontal axis, it meets the horizontal axis at the tick one unit left of the vertical axis and again at the tick three units right of it, and its two branches fall to the bottom edge of the panel, each ending in an arrow. Panel C shows part of a curve that opens downward; its highest point sits on the horizontal axis, two units left of the vertical axis, and its two branches fall to the bottom edge of the panel, each ending in an arrow. No equations, coordinate pairs or marked points are printed.
1. Hint 1
  A root is an $x$ -coordinate where the curve meets the horizontal axis.
2. Hint 2
  Count a touching point once, and count distinct crossing points separately.
Answer

A: $0$ roots; B: $2$ roots; C: $1$ root.

Full solution

Graph A lies entirely above the $x$ -axis, so no value of $x$ makes $y$ zero and its equation has no real roots.
Graph B crosses the axis in two places, at two different values of $x$ , so its equation has two real roots.
Graph C touches the axis at exactly one point, and a touching point counts once, so its equation has one real root.

Answer

A: $0$ roots; B: $2$ roots; C: $1$ root.

Key idea

The number of real roots is the number of distinct points where a parabola meets the horizontal axis.
Problem 6: Mark complete
Problem 7 Two requirements

Find every real value of $x$ that satisfies both $(x-4)(x+7)=0$ and $(2x-8)(x-9)=0$ .
1. Hint 1
  A value that satisfies both equations has to appear among the roots of each one.
2. Hint 2
  Set each factor equal to zero, then compare the two lists of roots you obtain.
Answer

$x=4$ , and no other value.

Full solution

The first equation gives $x=4$ or $x=-7$ .
In the second equation, $2x-8=0$ gives $x=4$ , and $x-9=0$ gives $x=9$ , so its roots are $4$ and $9$ .
Only one value appears in both lists.
$x=4$
That value makes the factor $x-4$ zero in the first equation and the factor $2x-8$ zero in the second, so both products are zero.
Substituting $-7$ in the second equation gives $(-22)(-16)=352$ , and substituting $9$ in the first gives $(5)(16)=80$ , so neither of those is a second common value.

Answer

$x=4$ , and no other value.

Key idea

A value asked to satisfy several equations at once must lie in the overlap of their root lists.
Problem 7: Mark complete
Problem 8 A walker on a marked trail

A straight trail is marked in kilometers and runs from the $18$ km point to the $35$ km point. A ranger station stands at the $20$ km point, and a walker on the trail is at the $p$ km point. The square of her distance in kilometers from the station is $64$ . Write the equation this gives, find every value of $p$ that satisfies it, and state which value the range of the trail rules out.
1. Hint 1
  The distance between two marked points is the gap between their numbers, and squaring that gap loses track of which of the two points lies farther along.
2. Hint 2
  Take square roots of both sides, keeping both signs, then add $20$ to each result and compare it with the range of the trail.
Answer

$(p-20)^2=64$ , giving $p=28$ or $p=12$ ; the range of the trail rules out $p=12$ , leaving $p=28$ .

Full solution

Her distance from the station is the gap between $p$ and $20$ , and squaring that gap gives $64$ .
$(p-20)^2=64$
Take square roots of both sides, keeping both signs.
$p-20=\pm8$
The positive choice gives $p=28$ , and the negative choice gives $p=12$ .
Both values satisfy the equation, since $8^2$ and $(-8)^2$ are both $64$ .
The trail starts at the $18$ km point, so the $12$ km point lies off it, while $28$ sits between $18$ and $35$ .
The walker is at the $28$ km point.

Answer

$(p-20)^2=64$ , giving $p=28$ or $p=12$ ; the range of the trail rules out $p=12$ , leaving $p=28$ .

Key idea

Both signs belong to the algebra, and the context decides which of the values it allows.
Problem 8: Mark complete
Problem 9 A sign prediction

A student says that the two real solutions of $(x-8)^2=9$ must have opposite signs. Decide whether this is correct and explain the role of the two square-root signs.
1. Hint 1
  Solve the equation first, then test the claim against the two values you get.
2. Hint 2
  First find the two values of $x-8$ , then add $8$ to each.
Answer

No; the solutions are $x=11$ and $x=5$ .

Full solution

The square-root step gives
$x-8=\pm3.$
Adding $8$ produces $x=11$ and $x=5$ .
Both are positive.
The expressions $11-8$ and $5-8$ are opposite, but shifting both by $8$ means the original values need not be opposite.

Answer

No; the solutions are $x=11$ and $x=5$ .

Key idea

The two square-root signs apply to the squared group, not necessarily to the final values of the variable.
Problem 9: Mark complete
Problem 10 A graph conclusion

The parabola $y=ax^2+bx+c$ has no point on the $x$ -axis. Sam concludes that $ax^2+bx+c=0$ has no real roots. Lee concludes that $ax^2+bx+c=0$ is not a quadratic equation. Assess both conclusions and explain.
1. Hint 1
  Separate the definition of a root from the definition of a quadratic equation.
2. Hint 2
  Try to build a parabola with no point on the $x$ -axis, then look at its squared-term coefficient.
Answer

Sam is correct; Lee is incorrect.

Full solution

No point on the $x$ -axis means no real value of $x$ gives $y=0$ , so the equation has no real root and Sam is correct.
The graph is a parabola, and if $a$ were $0$ the graph would be the straight line $y=bx+c$ rather than a U-shaped curve, so its squared-term coefficient is not zero, and $ax^2+bx+c=0$ is a quadratic equation however many real roots it has, which makes Lee wrong.
A concrete case is the parabola
$y=x^2+1.$
Every real $x$ gives $y\ge1$ , so this parabola never meets the axis, and its equation $x^2+1=0$ would need $x^2=-1$ , which no real number satisfies.
Its squared-term coefficient is $a=1$ , which is not zero, so a quadratic equation with no real root really does exist.

Answer

Sam is correct; Lee is incorrect.

Key idea

A quadratic equation may have no real roots while still having a nonzero squared-term coefficient.
Problem 10: Mark complete

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