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Quadratic Equations by Factoring

Sep 16, 2022
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Key Concepts

  • Solve quadratic equations by factoring
  • Zero product property or null factor law
  • Standard form of quadratic equation.

Introduction 

A quadratic equation of the form ax2 + bx + c = 0 can also be represented as  y = ax2 + bx + c.

The graph y = ax2 + bx + c, is in the form of a parabola used to locate the solutions or roots of the quadratic equation. 

Solving Quadratic Equation by Factoring 

The standard form of the quadratic equation is ax2 + bx+c, which forms a parabola. 

Zero-Product Property 

Zero product property says that if the product of two numbers is zero, then either of the numbers or both the numbers must be equal to zero.  

Example 1: 

parallel

Determine the solutions of equation (x−8)(9x−4)=0 by factoring.  

Solution: 

Step 1: Given equation is (x−8)(9x−4)=0  … (1) 

Step 2: Now we will use zero product property for equation (1), 

(x – 8) =0 or (9x – 4) =0 

parallel

x = 8 and x =4/9

The solutions of the quadratic equation (x−8)(9x−4)=0 are x = 8 and x =4/9.

Example 2: 

Determine the solutions of the quadratic equation x2+3x−4=0  by factoring. Give answers to 1 decimal place where appropriate. 

Solution: 

Step 1: Given equation is x2+3x−4=0, which is in the standard form. 

Step 2: Now find the set of factors to factorize x2+3x−4=0… (1) 

Set of factors: (1, – 4), (-1, 4), (2, -2) 

The factor set (-1, 4) gives the sum as 3 and the product as -4. 

So, we consider this set. 

Step 3: Now we will rewrite the standard form into factorized form. 

(x-1) (x+4) = 0 … (2) 

Step 4: Now we will use zero product property for equation (2), 

(x -1) =0 or (x+4) =0 

x = 1 and x = -4 

The solutions of the quadratic equation x2+3x−4=0  are x = 1 and x = -4. 

Plot the graph for the factorized quadratic equations 

We can plot the points and draw the graph for the quadratic equations. 

First, we will find the exact factors of the quadratic equations, then using these x-intercept values; we will find the vertex and then plot the graph. 

Let us learn this from an example. 

Example: 

Determine the solutions of the quadratic equation x2−7x+12=0  by factoring and plot the graph.  

Solution: 

Step 1: Given quadratic equation x2−7x+12=0  is in the standard form. 

Step 2: Now find the set of factors to factorize x2−7x+12=0… (1) 

Set of factors: (-4, – 3), (-6, 2), (2, -6) (12, -1) (-12, 1) 

The factor set (-4, -3) gives the sum as -7 and product as 12 

So, we consider this set. 

Step 3: Now we will rewrite the standard form into factorized form. 

(x – 4) (x – 3) = 0 … (2) 

Step 4: Now we will use zero product property for equation (2), 

(x – 4) =0 or (x – 3) = 0  

x = 4, x = 3 

The solutions of the quadratic equation x2−7x+12=0  are x = 4 and x = 3. 

Step 5: Now find the coordinates of vertex by taking the average of x-intercepts = (3+4)/2 = 3.5 

Now x coordinate is 3.5. 

Now substitute the x-coordinate in x2−7x+12=0 . 

We get, f(x)=3.52−7(3.5)+12= −0.25.

The vertex of the graph is (3.5, -0.25) 

Step 6: 

Plot the vertex and x-intercepts on the graph and draw the parabola. We get, 

Factored form of quadratic equation by the graph. 

Can we find the factored form of the quadratic equation using the graph? 

Let us see this. 

Example 1: 

Given a graph as shown in the image below, find the factored form of the quadratic equation. 

Example 1: 

Solution: 

Step 1: 

Now see the x-intercepts and note them from the graph. 

From the graph x=3, x=8 

Step 2: 

We write the quadratic equation in factored form: 

a(x – p)(x – q)=0

Substitute x-intercepts in the equation, we get 

a(x−3)(x−8)=0

Step 3: 

Now we find ‘a’ value by vertex (5.5, -6.25) 

f(x) = a(x−3)(x−8)

−6.25=a(5.5−3)(5.5−8)

−6.25=a(2.5)(−2.5)

a=1

So, the factorized form of the quadratic equation is 1(x – 3)(x – 8)=0.  

Example 2: 

Given a graph as shown in the image below, find the factored form of the quadratic equation. 

Example 2: 

Solution: 

Step 1: 

Now see the x-intercepts and note them from the graph. 

From the graph x=1, x=1 

Step 2: 

We write the quadratic equation in factored form: 

a(x – 1)(x – 1)=0

Substitute x-intercepts in the equation, we get 

a(x−1)(x−1)=0

Step 3: Write the quadratic equation 

The factorized form of the quadratic equation is

(x – 1)(x – 1)=0

The quadratic equation is 𝒙𝟐−𝟐𝒙+𝟏=𝟎 

Example 3: 

Given a graph as shown in the image below, find the factored form of the quadratic equation. 

Example 3: 

Solution: 

Step 1: 

Now see the x-intercepts and note them from the graph. 

From the graph x=3, x=3 

Step 2: 

We write the quadratic equation in factored form: 

a(x – P)(x – q)=0

Substitute x-intercepts in the equation, we get 

a(x−3)(x−3)=0

Step 3: Write the quadratic equation 

The factorized form of the quadratic equation is (x – 3)(x – 3)=0.  

x2−6x+9=0 is the quadratic equation. 

Verifying Solutions 

  • How can we verify whether the solution obtained is true or not? 

Yes, we can verify the solution by substituting the roots in the equation. 

Let us understand this from an example. 

Example: 

The solutions of the quadratic equation x2−7x+12=0 by factoring are x = 3 and x = 4.  

Solution:  

Let us substitute these values in the equation x2−7x+12=0 … (1). 

Let us take x = 3. 

We get,

L.H.S=(3)2−7(3)+12, 

L.H.S=9−21+12,

L.H.S=0=R.H.S  

When x = 4 

We get,

L.H.S=(4)2−7(4)+12, 

L.H.S=16−28+12,

L.H.S=0=R.H.S 

Solution is verified. 

Quadratic equations in Real Life 

Let us learn the use of quadratic equations in our real life. 

Example: 

The area of the image shown below is 196 sq.m; write the area in the form of the quadratic equation. 

Quadratic equations in Real Life 

Solution: 

From the image, we can write as (x+8)(x+18) = area 

Given area = 196 sq. m 

(x+8)(x+18)=196

x2+8x+18x+144 = 196

x2 +26x −52=0

This is the quadratic equation for the given image. 

Exercise

  1. Determine the solutions of the quadratic equation x²-10x+25= 0 by factoring.
  2. Determine the solutions of the quadratic equation x²-6x+5=0 by factoring.
  3. Determine the solutions of the quadratic equation x²-9x+9=0 by factoring.
  4. Convert the quadratic equation x² – 13x+12= 0 into factored form.
  5. Write the equation (x-3)(x-4)= 0 into the standard form of the quadratic equation.
  6. Determine the solutions of the quadratic equation 4x2-3x+10= 0 by factoring.
  7. Determine the solutions of the quadratic equation 3x2-2x+1 = 0 by factoring.
  8. Find the solution of the quadratic equation x² – 7x+12= 0 using zero product property.
  9. Determine the solutions of the quadratic equation x²-18x+72 = 0 by factoring.
  10. Determine the type of solutions for the quadratic equation x² + 15x+56= 0 by factoring.

Concept Map

Concept Map

What have we learned

  • Quadratic equations by factoring
  • Zero product property or null factor law
  • Standard form of quadratic equation

Comments:

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