Quadratic Formula Practice Problems - Step-by-Step Solutions

Master the quadratic formula with guided practice problems. Learn to solve ax²+bx+c=0 equations, identify coefficients, and understand discriminants through examples.

📚What You'll Master in This Practice Session
  • Identify coefficients a, b, and c in quadratic equations
  • Apply the quadratic formula X = (-b ± √(b²-4ac)) / 2a correctly
  • Calculate discriminant values to predict number of solutions
  • Solve complete quadratic equations with three parameters
  • Interpret parabola intersection points on coordinate axes
  • Check solutions by substituting back into original equations

Understanding The Quadratic Formula

Complete explanation with examples

What is a quadratic equation?

Quadratic equations (also called second degree equations) contain three numbers called parameters:

  • Parameter a a represents the position of the vertex of the parabola on the Y Y axis. A parabola can have a maximum vertex, or a minimum vertex (depending on if the parabola opens upwards or downwards).
  • Parameter b b represents the position of the vertex of the parabola on the X X axis.
  • Parameter c c represents the point of intersection of the parabola with the Y Y axis.

These three parameters are expressed in quadratic equations as follows:

aX2+bX+c=0 aX^2+bX+c=0

They are called the coefficients of the equation.

So, how do we find the value of X X ?

To find X X and be able to solve the quadratic equation, all we need to do is to input the parameters (the number values of a, b and c) from the equation into the quadratic formula, and solve for X X .

For example:

3X2+8X+4=0 3X^2+8X+4=0

Detailed explanation

Practice The Quadratic Formula

Test your knowledge with 25 quizzes

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number


\( 10x^2+5+20x=0 \)

What are the components of the equation?

Examples with solutions for The Quadratic Formula

Step-by-step solutions included
Exercise #1

What is the value of X in the following equation?

X2+10X+9=0 X^2+10X+9=0

Step-by-Step Solution

To answer the question, we'll need to recall the quadratic formula:

x=b±b24ac2a x = {-b \pm \sqrt{b^2-4ac} \over 2a}

 

Let's remember that:

a is the coefficient of X²

b is the coefficient of X

c is the free term

 

And if we look again at the formula given to us:

a=1

b=10

c=9

 

Let's substitute into the formula:

x=10±10241921 x = {-10 \pm \sqrt{10^2-4\cdot 1 \cdot 9} \over 2\cdot 1}

Let's start by solving what's under the square root:

x=10±100362 x = {-10 \pm \sqrt{100-36} \over 2}

x=10±642 x = {-10 \pm \sqrt{64} \over 2}

x=10±82 x = {-10 \pm 8 \over 2}

Now we'll solve twice, once with plus and once with minus

 

x=10+82=22=1 x = {-10 +8 \over 2}= {-2 \over 2} = -1

x=1082=182=9 x = {-10 -8 \over 2} = {-18 \over 2} =-9

And we can see that we got two solutions, X=-1 and X=-9

And that's the solution!

Answer:

x1=1,x2=9 x_1=-1,x_2=-9

Video Solution
Exercise #2

Solve for X:

2X2+6X+8=0 -2X^2+6X+8=0

Step-by-Step Solution

To solve the quadratic equation 2X2+6X+8=0 -2X^2 + 6X + 8 = 0 using the quadratic formula, follow these steps:

  • Step 1: Calculate the discriminant, b24ac b^2 - 4ac .

Here, a=2 a = -2 , b=6 b = 6 , and c=8 c = 8 . Plug these into the formula: b24ac=624(2)(8)=36+64=100 b^2 - 4ac = 6^2 - 4(-2)(8) = 36 + 64 = 100 Since the discriminant is greater than zero, the roots are real and distinct.

  • Step 2: Apply the quadratic formula, X=b±b24ac2a X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} .

Substituting the values, we have: X=6±1002(2) X = \frac{-6 \pm \sqrt{100}}{2(-2)} Simplifying inside the square root gives us: X=6±104 X = \frac{-6 \pm 10}{-4} This leads to two possible solutions: - First, calculate with the positive square root: X1=6+104=44=1 X_1 = \frac{-6 + 10}{-4} = \frac{4}{-4} = -1 - Second, calculate with the negative square root: X2=6104=164=4 X_2 = \frac{-6 - 10}{-4} = \frac{-16}{-4} = 4

Thus, the solutions to the equation are X1=4 X_1 = 4 and X2=1 X_2 = -1 .

Verifying against the choices, the correct choice is: : X1=4,X2=1 X_1=4, X_2=-1 .

Therefore, the solution is X1=4,X2=1 X_1 = 4, X_2 = -1 .

Answer:

X1=4,X2=1 X_1=4, X_2=-1

Video Solution
Exercise #3

Solve the following equation:

x2+5x+4=0 x^2+5x+4=0

Step-by-Step Solution

The parameters are expressed in the quadratic equation as follows:

aX2+bX+c=0

 

We substitute into the formula:

 

-5±√(5²-4*1*4) 
          2

 

-5±√(25-16)
         2

 

-5±√9
    2

 

-5±3
   2

 

The symbol ± means that we have to solve this part twice, once with a plus and a second time with a minus,

This is how we later get two results.

 

-5-3 = -8
-8/2 = -4

 

-5+3 = -2
-2/2 = -1

 

And thus we find out that X = -1, -4

Answer:

x1=1 x_1=-1 x2=4 x_2=-4

Video Solution
Exercise #4

Solve the following equation:

x2+9x+8=0 x^2+9x+8=0

Step-by-Step Solution

To solve the quadratic equation x2+9x+8=0 x^2 + 9x + 8 = 0 , we will use the factoring method because it appears simple to factor.

First, we attempt to factor the quadratic expression x2+9x+8 x^2 + 9x + 8 . We look for two numbers that multiply to 8 (the constant term) and add up to 9 (the coefficient of the x x term).

These numbers are 1 and 8. So, we can write:

x2+9x+8=(x+1)(x+8)=0 x^2 + 9x + 8 = (x + 1)(x + 8) = 0

Now, to find the solutions, we set each factor equal to zero:

  1. x+1=0 x + 1 = 0 x=1 x = -1
  2. x+8=0 x + 8 = 0 x=8 x = -8

Therefore, the solutions to the equation are x1=1 x_1 = -1 and x2=8 x_2 = -8 .

Upon reviewing the multiple-choice answers, we find that the correct choice is the one that matches our solutions:

x1=1 x_1=-1 x2=8 x_2=-8

Answer:

x1=1 x_1=-1 x2=8 x_2=-8

Video Solution
Exercise #5

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

Frequently Asked Questions

What is the quadratic formula and when do I use it?

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The quadratic formula is X = (-b ± √(b²-4ac)) / 2a, used to solve second-degree equations in the form ax²+bx+c=0. Use it when you can't easily factor the quadratic equation or when completing the square becomes too complex.

How do I identify the coefficients a, b, and c in a quadratic equation?

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In the standard form ax²+bx+c=0: 'a' is the coefficient of X², 'b' is the coefficient of X, and 'c' is the constant term. For example, in 3X²+8X+4=0, a=3, b=8, c=4. Remember that 'a' cannot equal zero.

What does the discriminant tell me about solutions?

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The discriminant (b²-4ac) predicts the number of solutions: If positive (>0), there are two real solutions; if zero (=0), there's one solution; if negative (<0), there are no real solutions. This corresponds to how many times the parabola crosses the X-axis.

Why do quadratic equations sometimes have two solutions?

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Quadratic equations represent parabolas, which are U-shaped curves. A parabola can intersect the X-axis at two points, one point, or no points at all. Each intersection point represents a solution to the equation, which is why you can have zero, one, or two solutions.

What's the difference between quadratic equations and linear equations?

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Linear equations have variables to the first power (like 2x+3=0) and graph as straight lines, while quadratic equations have variables to the second power (like x²+2x+1=0) and graph as curved parabolas. Quadratic equations can have up to two solutions, while linear equations have exactly one solution.

How do I solve quadratic equations step by step?

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Follow these steps: 1) Write the equation in standard form ax²+bx+c=0, 2) Identify coefficients a, b, and c, 3) Substitute into the quadratic formula X = (-b ± √(b²-4ac)) / 2a, 4) Calculate the discriminant under the square root, 5) Solve for both + and - cases, 6) Check your answers by substituting back.

Can I use the quadratic formula for incomplete quadratic equations?

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Yes, but you need to identify missing terms correctly. For equations like X²+5X=0, you have a=1, b=5, c=0. For equations like X²-4=0, you have a=1, b=0, c=-4. Always ensure you're working with the complete form ax²+bx+c=0 before applying the formula.

What should I do if I get a negative number under the square root?

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A negative discriminant means the equation has no real solutions - the parabola doesn't intersect the X-axis. Double-check your calculations first, but if correct, simply state 'no real solutions' or 'no solution' as your answer. This is a valid mathematical result.

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