Quadratic Equations Practice Problems - Solve by 3 Methods

Master solving quadratic equations with step-by-step practice problems using trinomial factoring, quadratic formula, and completing the square methods.

📚Master Three Essential Methods for Solving Quadratic Equations
  • Factor trinomials using the ac method to find quadratic solutions
  • Apply the quadratic formula to solve ax²+bx+c=0 equations systematically
  • Complete the square method for quadratic functions with any coefficients
  • Identify when to use each method based on coefficient patterns
  • Solve real-world problems involving quadratic equations and functions
  • Verify solutions by substitution and graphical interpretation

Understanding Solving Quadratic Equations

Complete explanation with examples

Methods for solving a quadratic function

In this article, we will learn the three most common ways to solve a quadratic function easily and quickly.

  1. Trinomial
  2. Quadratic Formula
  3. Completing the Square

Reminder:

The basic quadratic function equation is:
Y=ax2+bx+cY=ax^2+bx+c

When:
a a   - the coefficient of X2X^2
b b   - the coefficient of XX
cc - the constant term

  • aa must be different from 00
  • bb or cc can be 00
  • a,b,ca,b,c can be negative/positive
  • The quadratic function can also look like this:
    • Y=ax2Y=ax^2
    • Y=ax2+bxY=ax^2+bx
    • Y=ax2+cY=ax^2+c
Detailed explanation

Practice Solving Quadratic Equations

Test your knowledge with 28 quizzes

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number


what is the value of \( a \) in the equation

\( y=-x^2-3x+1 \)

Examples with solutions for Solving Quadratic Equations

Step-by-step solutions included
Exercise #1

Solve the following equation:

2x210x12=0 2x^2-10x-12=0

Step-by-Step Solution

Let's recall the quadratic formula:

Quadratic formula | The formula

We'll substitute the given data into the formula:

x=(10)±10242(12)22 x={{-(-10)\pm\sqrt{-10^2-4\cdot2\cdot(-12)}\over 2\cdot2}}

Let's simplify and solve the part under the square root:

x=10±100+964 x={{10\pm\sqrt{100+96}\over 4}}

x=10±1964 x={{10\pm\sqrt{196}\over 4}}

x=10±144 x={{10\pm14\over 4}}

Now we'll solve using both methods, once with the addition sign and once with the subtraction sign:

x=10+144=244=6 x={{10+14\over 4}} = {24\over4}=6

x=10144=44=1 x={{10-14\over 4}} = {-4\over4}=-1

We've arrived at the solution: X=6,-1

Answer:

x1=6 x_1=6 x2=1 x_2=-1

Video Solution
Exercise #2

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number


x2+7x=0 x^2+7x=0

What are the components of the equation?

Step-by-Step Solution

Let's solve the problem step-by-step:

First, consider the given equation:

x2+7x=0 x^2 + 7x = 0

This equation is almost in the standard form of a quadratic equation:

ax2+bx+c=0 ax^2 + bx + c = 0

Where:

  • a a is the coefficient of x2 x^2
  • b b is the coefficient of x x
  • c c is the constant term (independent number)

Let's identify each of these components from the given equation:

  • For a a : The term with x2 x^2 is x2 x^2 . In this case, the coefficient is implicitly 1, so a=1 a = 1 .
  • For b b : The term with x x is 7x 7x . The coefficient of x x is 7, so b=7 b = 7 .
  • For c c : There is no independent constant term visible, so we assume c=0 c = 0 .

Thus, the components of the quadratic equation are:

a=1 a = 1 , b=7 b = 7 , c=0 c = 0

The correct choice from the provided options is : a=1 a=1 , b=7 b=7 , c=0 c=0

Answer:

a=1 a=1 b=7 b=7 c=0 c=0

Video Solution
Exercise #3

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number


56x2+12x=0 5-6x^2+12x=0

What are the components of the equation?

Step-by-Step Solution

Let's solve this problem step-by-step by identifying the coefficients of the quadratic equation:

First, examine the given equation:

56x2+12x=05 - 6x^2 + 12x = 0

To make it easier to identify the coefficients, we rewrite the equation in the standard quadratic form:

6x2+12x+5=0-6x^2 + 12x + 5 = 0

In this expression, we can now directly identify the coefficients:

  • The coefficient of x2 x^2 (quadratic term) is a=6 a = -6 .
  • The coefficient of x x (linear term) is b=12 b = 12 .
  • The constant term (independent number) is c=5 c = 5 .

Thus, the components of the quadratic equation are:

a=6 a = -6 , b=12 b = 12 , c=5 c = 5

By comparing these values to the multiple-choice options, we can determine that the correct choice is:

Choice 4: a=6 a = -6 , b=12 b = 12 , c=5 c = 5

Therefore, the final solution is:

a=6 a = -6 , b=12 b = 12 , c=5 c = 5 .

Answer:

a=6 a=-6 b=12 b=12 c=5 c=5

Video Solution
Exercise #4

a = coefficient of x²

b = coefficient of x

c = coefficient of the constant term


What is the value of c c in the function y=x2+25x y=-x^2+25x ?

Step-by-Step Solution

Let's recall the general form of the quadratic function:

y=ax2+bx+c y=ax^2+bx+c The function given in the problem is:

y=x2+25x y=-x^2+25x c c is the free term (meaning the coefficient of the term with power 0),

In the function in the problem there is no free term,

Therefore, we can identify that:

c=0 c=0 Therefore, the correct answer is answer A.

Answer:

c=0 c=0

Video Solution
Exercise #5

Solve the following equation:

x2+3x18=0 x^2+3x-18=0

Step-by-Step Solution

This is a quadratic equation:

x2+3x18=0 x^2+3x-18=0

This is due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed to solve it using the quadratic formula,

Remember:

The rule states that the roots of the equation of the form:

ax2+bx+c=0 ax^2+bx+c=0

are:

x1,2=b±b24ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

This formula is called: "The Quadratic Formula"

Let's return to the problem:

x2+3x18=0 x^2+3x-18=0

And solve it:

First, let's identify the coefficients of the terms:

{a=1b=3c=18 \begin{cases}a=1\\b=3\\c=-18\end{cases}

where we noted that the coefficient of the quadratic term is 1,

And we'll obtain the solutions of the equation (its roots) by substituting the coefficients we just noted in the quadratic formula:

x1,2=b±b24ac2a=3±3241(18)21 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-3\pm\sqrt{3^2-4\cdot1\cdot(-18)}}{2\cdot1}

Let's continue and calculate the expression inside the square root and simplify the expression:

x1,2=3±812=3±92 x_{1,2}=\frac{-3\pm\sqrt{81}}{2}=\frac{-3\pm9}{2}

Therefore the solutions of the equation are:

{x1=3+92=3x2=392=6 \begin{cases}x_1=\frac{-3+9}{2}=3 \\ x_2=\frac{-3-9}{2}=-6\end{cases}

Therefore the correct answer is answer C.

Answer:

x1=3,x2=6 x_1=3,x_2=-6

Video Solution

Frequently Asked Questions

What are the three main methods for solving quadratic equations?

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The three main methods are: 1) Trinomial factoring (finding two numbers that multiply to ac and add to b), 2) Quadratic formula (x = [-b ± √(b²-4ac)]/2a), and 3) Completing the square (rewriting as a perfect square trinomial).

When should I use the trinomial method vs quadratic formula?

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Use trinomial factoring when a=1 and the equation factors easily with integer solutions. Use the quadratic formula when factoring is difficult or when you need exact decimal solutions. The quadratic formula works for all quadratic equations.

How do I know if a quadratic equation can be factored?

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Check if the discriminant (b²-4ac) is a perfect square. If it is, the equation factors with rational solutions. Also look for common factors first, and see if you can find two numbers that multiply to ac and add to b.

What is the discriminant and why is it important?

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The discriminant is b²-4ac from under the square root in the quadratic formula. If it's positive, there are two real solutions. If zero, one repeated solution. If negative, no real solutions (complex solutions only).

How do I complete the square for ax²+bx+c when a≠1?

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First factor out 'a' from the x terms: a(x² + (b/a)x) + c. Then complete the square inside parentheses by adding and subtracting (b/2a)². Finally, factor the perfect square trinomial and simplify.

What are common mistakes when solving quadratic equations?

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Common errors include: forgetting the ± in the quadratic formula, sign errors when completing the square, not checking if a=0 (making it linear), and forgetting to set the equation equal to zero before factoring.

Can quadratic equations have no real solutions?

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Yes, when the discriminant (b²-4ac) is negative, there are no real number solutions. The parabola doesn't cross the x-axis. However, there are still complex number solutions using imaginary numbers.

How do I check if my quadratic equation solutions are correct?

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Substitute each solution back into the original equation. If both sides equal zero, your solution is correct. You can also verify by graphing - the solutions are where the parabola crosses the x-axis.

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