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

Practice Solving Quadratic Equations

Examples with solutions for Solving Quadratic Equations

Exercise #1

Solve the following equation:

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

Video Solution

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

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?

Video Solution

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

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?

Video Solution

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

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 ?

Video Solution

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

Exercise #5

Solve the following equation:

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

Video Solution

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

Exercise #6

Solve the following equation:

x24x+4=0 x^2-4x+4=0

Video Solution

Step-by-Step Solution

The given equation is:

x24x+4=0 x^2 - 4x + 4 = 0

This resembles a perfect square trinomial. The expression x24x+4 x^2 - 4x + 4 can be rewritten as (x2)2 (x-2)^2 . This can be verified by expanding (x2)(x2) (x-2)(x-2) to confirm it equals x24x+4 x^2 - 4x + 4 .

Therefore, the equation becomes:

(x2)2=0 (x-2)^2 = 0

To solve for x x , take the square root of both sides:

x2=0 x - 2 = 0

Adding 2 to both sides gives:

x=2 x = 2

Thus, the solution to the equation x24x+4=0 x^2 - 4x + 4 = 0 is x=2 x = 2 , which corresponds to the unique real root of the equation.

Answer

x=2 x=2

Exercise #7

Solve the following equation:

x2+10x+25=0 x^2+10x+25=0

Video Solution

Step-by-Step Solution

The given quadratic equation is x2+10x+25=0 x^2 + 10x + 25 = 0 .

Notice that x2+10x+25 x^2 + 10x + 25 is a perfect square trinomial, which can be factored as (x+5)2 (x + 5)^2 . Let's verify by expanding:

(x+5)2=(x+5)(x+5)=x2+5x+5x+25=x2+10x+25(x + 5)^2 = (x + 5)(x + 5) = x^2 + 5x + 5x + 25 = x^2 + 10x + 25.

Since the factoring is correct, we rewrite the equation:

(x+5)2=0(x + 5)^2 = 0 .

Take the square root of both sides:

x+5=0x + 5 = 0 .

Solving for x x , we find:

x=5x = -5.

The equation has a double root, meaning the solution is repeated. Thus, the final solution is:

x=5x = -5.

This matches the given correct answer.

Answer

x=5 x=-5

Exercise #8

Solve the following equation:

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

Video Solution

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

Exercise #9

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number


x2+4x5=0 x^2+4x-5=0

What are the components of the equation?

Video Solution

Step-by-Step Solution

The quadratic equation we have is x2+4x5=0 x^2 + 4x - 5 = 0 .

We'll compare this with the general form of a quadratic equation: ax2+bx+c=0 ax^2 + bx + c = 0 .

1. Identify a a : The coefficient of x2 x^2 in the given equation is 1 1 . Therefore, a=1 a = 1 .

2. Identify b b : The coefficient of x x in the given equation is 4 4 . Therefore, b=4 b = 4 .

3. Identify c c : The constant term in the given equation is 5 -5 . Therefore, c=5 c = -5 .

Thus, the components of the equation are:

  • a=1 a = 1
  • b=4 b = 4
  • c=5 c = -5

The correct answer to this problem, matching choice id 3, is:

a=1 a=1 b=4 b=4 c=5 c=-5

Answer

a=1 a=1 b=4 b=4 c=5 c=-5

Exercise #10

Solve the following problem:

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

Video Solution

Step-by-Step Solution

This is a quadratic equation:

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

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 to a form where all terms on one side are ordered from the highest to the 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 an 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+5x+4=0=0 x^2+5x+4=0 =0

And solve it:

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

{a=1b=5c=4 \begin{cases}a=1 \\ b=5 \\ c=4\end{cases}

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

We obtain the solutions of the equation (its roots) by insertion we just identified into the quadratic formula:

x1,2=b±b24ac2a=5±5241421 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot4}}{2\cdot1}

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

x1,2=5±92=5±32 x_{1,2}=\frac{-5\pm\sqrt{9}}{2}=\frac{-5\pm3}{2}

Therefore the solutions of the equation are:

{x1=5+32=1x2=532=4 \begin{cases}x_1=\frac{-5+3}{2}=-1 \\ x_2=\frac{-5-3}{2}=-4\end{cases}

Therefore the correct answer is answer C.

Answer

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

Exercise #11

Solve the following equation:

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

Video Solution

Step-by-Step Solution

This is a quadratic equation:

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

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 to a form where all the terms on one side are ordered from the highest to the 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 an 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+5x+6=0 x^2+5x+6=0 and solve it:

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

{a=1b=5c=6 \begin{cases}a=1 \\ b=5 \\ c=6\end{cases}

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

We obtain the equation's solutions (roots) by inserting the coefficients we just noted into the quadratic formula:

x1,2=b±b24ac2a=5±5241621 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot6}}{2\cdot1}

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

x1,2=5±12=5±12 x_{1,2}=\frac{-5\pm\sqrt{1}}{2}=\frac{-5\pm1}{2}

The solutions to the equation are:

{x1=5+12=2x2=512=3 \begin{cases}x_1=\frac{-5+1}{2}=-2 \\ x_2=\frac{-5-1}{2}=-3\end{cases}

Therefore the correct answer is answer D.

Answer

x1=3,x2=2 x_1=-3,x_2=-2

Exercise #12

What is the value of X in the following equation?

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

Video Solution

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

Exercise #13

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number


what is the value of a a in the equation

y=x23x+1 y=-x^2-3x+1

Video Solution

Step-by-Step Solution

To determine the coefficient a a in the given quadratic equation y=x23x+1 y = -x^2 - 3x + 1 , follow these steps:

  • Step 1: Recognize the form of the quadratic equation as y=ax2+bx+c y = ax^2 + bx + c .
  • Step 2: Identify the x2 x^2 term in the equation y=x23x+1 y = -x^2 - 3x + 1 .
  • Step 3: Determine the coefficient of the x2 x^2 term, which is in front of x2 x^2 .

In the equation y=x23x+1 y = -x^2 - 3x + 1 , the term involving x2 x^2 is x2-x^2, where the coefficient a a is clearly 1-1.

Hence, the value of a a is a=1 a = -1 .

Answer

a=1 a=-1

Exercise #14

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number


10x2+5+20x=0 10x^2+5+20x=0

What are the components of the equation?

Video Solution

Step-by-Step Solution

To determine the components of the quadratic equation, follow these steps:

  • Step 1: Recognize the standard form of a quadratic equation, which is ax2+bx+c=0 ax^2 + bx + c = 0 .
  • Step 2: Compare the given equation 10x2+20x+5=0 10x^2 + 20x + 5 = 0 to the standard form.
  • Step 3: Identify the coefficients:
    - The term 10x2 10x^2 indicates that a=10 a = 10 .
    - The term 20x 20x indicates that b=20 b = 20 .
    - The term 5 5 is the constant term, so c=5 c = 5 .

Therefore, the components of the equation are:

a=10 a = 10 , b=20 b = 20 , c=5 c = 5 .

The correct answer among the choices provided is the one that correctly identifies these coefficients:

a=10 a=10 b=20 b=20 c=5 c=5

Therefore, the correct choice is Choice 4.

Answer

a=10 a=10 b=20 b=20 c=5 c=5

Exercise #15

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

Identifies a,b,c

5x2+6x8=0 5x^2+6x-8=0

Video Solution

Step-by-Step Solution

To identify the coefficients from the quadratic equation 5x2+6x8=0 5x^2 + 6x - 8 = 0 , follow these steps:

  • Standard Form of Quadratic Equation: Compare 5x2+6x85x^2 + 6x - 8 with the standard form ax2+bx+c=0ax^2 + bx + c = 0.
  • Identify aa, bb, and cc:
    • a=5 a = 5 : The coefficient of x2x^2.
    • b=6 b = 6 : The coefficient of xx.
    • c=8 c = -8 : The constant term (independent of xx).

Therefore, from the equation 5x2+6x8=0 5x^2 + 6x - 8 = 0 , the coefficients are identified as a=5 a=5 , b=6 b=6 , and c=8 c=-8 .

Comparing with choices, we find that choice 2 is correct: a=5 a=5 , b=6 b=6 , c=8 c=-8 .

Thus, the coefficients are identified as a=5 a=5 , b=6 b=6 , c=8 c=-8 .

Answer

a=5 a=5 b=6 b=6 c=8 c=-8