Quadratic Equations: Practice Worksheet

Factorisation, the quadratic formula, discriminants, roots, and application problems.

18 problems·35–45 min·★★★★★

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Problem 1
Quadratic Equations·★★☆☆☆
Which equation is not quadratic after simplification?
1. (A)$x^3 + x = (x + 1)^2$
2. (B)$x(x + 2) = 3x - 2$
3. (C)$(x - 1)^2 + 2 = 3x$
4. (D)$(2x + 1)(x - 2) = x + 4$
▶Answer
Option A: $x^3 + x = (x + 1)^2$
▶Step-by-step solution
Expand both sides and bring every term to one side.
Its simplified form does not have degree two.
Therefore, option A is the required equation.
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Problem 2
Factorisation·★★☆☆☆
Solve $2x^2 - 4x - 16 = 0$ using factorisation.
▶Answer
The roots are $x = -2$ or $x = 4$.
▶Step-by-step solution
Clear any fractional coefficients, then split the middle term and factorise.
Set each linear factor equal to zero.
Hence, $x = -2$ or $x = 4$.
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Problem 3
Nature of Roots·★★★☆☆
Determine the nature of the roots of $x^2 - 16 = 0$
1. (A)no real roots
2. (B)two equal real roots
3. (C)two distinct real roots
4. (D)more than two real roots
▶Answer
The equation has two distinct real roots (option C).
▶Step-by-step solution
Calculate $D=b^2-4ac=64$.
$D>0$ gives two distinct real roots, $D=0$ gives equal roots, and $D<0$ gives no real roots.
Therefore, this equation has two distinct real roots.
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Problem 4
Quadratic Word Problems·★★★★☆
If a person's age were 6 years less, the square of that age would be 4 more than 6 times the present age. Find the present age.
▶Answer
The present age is $16$ years.
▶Step-by-step solution
Let the present age be $a$. Then $(a-6)^2=6a+4$.
The resulting quadratic has roots $16$ and $2$.
Reject the non-positive value $2$; the present age is $16$ years.
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Problem 5
Quadratic Word Problems·★★★★★
A train covers 330 km at a constant speed. At 5 km/h faster, it would take 55 minutes less. Find its original speed.
▶Answer
The original speed is $40$ km/h.
▶Step-by-step solution
Let the original speed be $v$ km/h. The time difference is $330/v-330/(v+5)=\frac{11}{12}$ hour.
Clearing denominators gives a quadratic whose roots are $40$ and $-45$.
A speed must be positive, so the original speed is $40$ km/h.
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Problem 6
Quadratic Word Problems·★★★★★
A parent's present age is 2 more than the square of the child's present age. When the child reaches the parent's present age, the parent will be 4 more than 23 times the child's present age. Find both present ages.
▶Answer
The child is $12$ years old and the parent is $146$ years old.
▶Step-by-step solution
Let the child's age be $c$; then the parent's age is $c^2+2$.
When the child reaches that age, the parent's age will be $2(c^2+2)-c$. Use the stated comparison to form the quadratic.
The valid age is $c=12$, giving parent age $12^2+2=146$ years.
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Problem 7
Roots of Quadratic Equations·★★☆☆☆
Which equation has $-3$ as a root?
1. (A)$3x^2 - 9x - 54 = 0$
2. (B)$3x^2 - 9x - 56 = 0$
3. (C)$3x^2 - 9x - 51 = 0$
4. (D)$3x^2 - 9x - 53 = 0$
▶Answer
Option A: $3x^2 - 9x - 54 = 0$
▶Step-by-step solution
Substitute $x = -3$ into each equation.
For option A, the left side becomes $0$.
The other options give non-zero values, so they do not have the stated root.
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Problem 8
Roots of Quadratic Equations·★★★☆☆
Is $-1$ a root of $x^2 - 1 = 0$? Justify your answer.
▶Answer
Yes.
▶Step-by-step solution
Substitute $x = -1$ into the left side.
The resulting value is $0$.
Since the value is $0$, the number is a root.
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Problem 9
Roots of Quadratic Equations·★★★☆☆
If $1$ is a root of $3x^2 + kx - 10 = 0$, find $k$.
1. (A)$-7$
2. (B)$7$
3. (C)$6$
4. (D)$8$
▶Answer
$k = 7$ (option B).
▶Step-by-step solution
Substitute $x = 1$ in the equation.
This gives $3 + 1k + -10 = 0$.
Solving the linear equation gives $k = 7$.
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Problem 10
Relations Between Roots·★★★☆☆
Which equation has sum of roots equal to $1$?
1. (A)$3x^2 - 3x - 12 = 0$
2. (B)$2x^2 - 6x - 4 = 0$
3. (C)$2x^2 - 4x - 6 = 0$
4. (D)$x^2 - 5 = 0$
▶Answer
Option A: $3x^2 - 3x - 12 = 0$
▶Step-by-step solution
For $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$.
For option A, $-b/a = 1$.
The coefficient ratios in the other options give different sums.
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Problem 11
Nature of Roots·★★★★☆
Which equation has two distinct real roots?
1. (A)$x^2 + 4x - 5 = 0$
2. (B)$x^2 + 10x + 25 = 0$
3. (C)$x^2 + 12x + 36 = 0$
4. (D)$x^2 + 6x + 10 = 0$
▶Answer
Option A: $x^2 + 4x - 5 = 0$
▶Step-by-step solution
Compute $D=b^2-4ac$ for each option.
For option A, $D=36$.
This discriminant corresponds to two distinct real roots.
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Problem 12
Nature of Roots·★★★★☆
Which equation has no real roots?
1. (A)$3x^2 - 7x + 6 = 0$
2. (B)$x^2 - 16x + 64 = 0$
3. (C)$x^2 - 8x + 12 = 0$
4. (D)$x^2 - 14x + 49 = 0$
▶Answer
Option A: $3x^2 - 7x + 6 = 0$
▶Step-by-step solution
Compute $D=b^2-4ac$ for each option.
For option A, $D=-23$.
This discriminant corresponds to no real roots.
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Problem 13
Nature of Roots·★★★★☆
For which values of $k$ does $x^2-kx+25=0$ have equal roots?
1. (A)$k=-10$ or $k=10$
2. (B)$k=-10$ only
3. (C)$k=25$
4. (D)$k=10$ only
▶Answer
$k=-10$ or $k=10$ (option A).
▶Step-by-step solution
Equal roots require the discriminant to be zero.
Here, $D=(-k)^2-4(25)=k^2-100$.
Solving $D=0$ gives $k=-10$ or $k=10$.
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Problem 14
Nature of Roots·★★★★☆
Does $x^2 - 4x + 5 = 0$ have two distinct real roots? Justify without solving for the roots.
▶Answer
No.
▶Step-by-step solution
Its discriminant is $D=b^2-4ac=-4$.
Two distinct real roots require $D>0$.
Here $D$ is negative, so the answer is no.
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Problem 15
Nature of Roots·★★★★★
True or false? Every quadratic equation has at most two distinct real roots. Justify your answer.
▶Answer
True.
▶Step-by-step solution
Use the discriminant or a counterexample to test the statement.
A non-zero polynomial of degree two cannot have more than two distinct roots.
Therefore, the statement is true.
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Problem 16
Relations Between Roots·★★★★★
For $x^2 + bx + c = 0$, suppose $b=0$ and $c=-8$. Must the two roots be equal in magnitude and opposite in sign? Explain.
▶Answer
Yes. The roots are equal in magnitude and opposite in sign.
▶Step-by-step solution
The equation becomes $x^2 - 8=0$.
Thus $x^2=8$, so $x=\sqrt{8}$ or $x=-\sqrt{8}$.
The two values have the same magnitude and opposite signs.
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Problem 17
Completing the Square·★★☆☆☆
When completing the square directly in $x^2 +6x$, which constant must be added and subtracted?
1. (A)$9$
2. (B)$3$
3. (C)$12$
4. (D)$11$
▶Answer
Add and subtract $9$ (option A).
▶Step-by-step solution
Write $1x^2 +6x$ as the beginning of $(1x + m)^2$.
Matching the middle term gives $m=3$.
Therefore the required constant is $m^2=9$.
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Problem 18
Quadratic Formula·★★★☆☆
Solve $2x^2 + 18x + 40 = 0$ using the quadratic formula.
▶Answer
The roots are $x = -5$ or $x = -4$.
▶Step-by-step solution
Use $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=2$, $b=18$, and $c=40$.
The discriminant is $D=4$.
Hence, $x = -5$ or $x = -4$.
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