Solving Equations by Factoring: Solving the problem

Examples with solutions for Solving Equations by Factoring: Solving the problem

Exercise #1

Solve for x.

x27x12=0 -x^2-7x-12=0

Video Solution

Step-by-Step Solution

First, factor using trinomials and remember that there might be more than one solution for the value of x x :

x27x12=0 -x^2-7x-12=0

Divide by -1:

x2+7x+12=0 x^2+7x+12=0

(x+3)(x+4)=0 (x+3)(x+4)=0

Therefore:

x+4=0 x+4=0

x=4 x=-4

Or:

x+3=0 x+3=0

x=3 x=-3

Answer

x=3,x=4 x=-3,x=-4

Exercise #2

Find the value of the parameter x.

2x27x+5=0 2x^2-7x+5=0

Video Solution

Step-by-Step Solution

We will factor using trinomials, remembering that there is more than one solution for the value of X:

2x27x+5=0 2x^2-7x+5=0

We will factor -7X into two numbers whose product is 10:

2x25x2x+5=0 2x^2-5x-2x+5=0

We will factor out a common factor:

2x(x1)5(x1)=0 2x(x-1)-5(x-1)=0

(2x5)(x1)=0 (2x-5)(x-1)=0

Therefore:

x1=0 x-1=0 x=1 x=1

Or:

2x5=0 2x-5=0

2x=5 2x=5

x=2.5 x=2.5

Answer

x=1,x=2.5 x=1,x=2.5

Exercise #3

Find the value of the parameter x.

x225=0 x^2-25=0

Video Solution

Step-by-Step Solution

We will factor using the shortened multiplication formulas:

a2b2=(ab)(a+b) a^2-b^2=(a-b)(a+b) (ab)2=a22ab+b2 (a-b)^2=a^2-2ab+b^2

(a+b)2=a2+2ab+b2 (a+b)^2=a^2+2ab+b^2

Remember that there might be more than one solution for the value of x.

According to the first formula:

x2=a2 x^2=a^2

We'll take the square root:

x=a x=a

25=b2 25=b^2

We'll take the square root:

b=5 b=5

We'll use the first shortened multiplication formula:

a2b2=(ab)(a+b) a^2-b^2=(a-b)(a+b)

x225=(x5)(x+5)=0 x^2-25=(x-5)(x+5)=0

Therefore:

x+5=0 x+5=0

x=5 x=-5

Or:

x5=0 x-5=0

x=5 x=5

Answer

x=5,x=5 x=5,x=-5

Exercise #4

Find the value of the parameter x.

(x5)2=0 (x-5)^2=0

Video Solution

Step-by-Step Solution

We will factor using the shortened multiplication formulas:

a2b2=(ab)(a+b) a^2-b^2=(a-b)(a+b) (ab)2=a22ab+b2 (a-b)^2=a^2-2ab+b^2

(a+b)2=a2+2ab+b2 (a+b)^2=a^2+2ab+b^2

Let's remember that there might be more than one solution for the value of x.

According to one solution, we'll take the square root:

(x5)2=0 (x-5)^2=0

x5=0 x-5=0

x=5 x=5

According to the second solution, we'll use the shortened multiplication formula:

(x5)2=x210x+25=0 (x-5)^2=x^2-10x+25=0

We'll use the trinomial:

(x5)(x5)=0 (x-5)(x-5)=0

x5=0 x-5=0

x=5 x=5

or

x5=0 x-5=0

x=5 x=5

Therefore, according to all calculations, x=5 x=5

Answer

x=5 x=5

Exercise #5

Find the value of the parameter x.

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

Video Solution

Step-by-Step Solution

To solve the equation (x+5)2=0(x+5)^2 = 0, we will use the fact that a perfect square is zero only when the quantity being squared is zero itself.

  • Step 1: Set x+5=0 x+5 = 0 since it is the term being squared.
  • Step 2: Solve for x x by subtracting 5 from both sides: x+5=0 x+5 = 0 x=5 x = -5

Therefore, the solution to the equation is x=5 x = -5 .

Answer

x=5 x=-5

Exercise #6

Find the value of the parameter x.

12x39x23x=0 12x^3-9x^2-3x=0

Video Solution

Step-by-Step Solution

To solve the problem, we follow these steps:

  • Step 1: Identify the given cubic equation, 12x39x23x=0 12x^3 - 9x^2 - 3x = 0 .
  • Step 2: Look for common factors in the terms of the equation.
  • Step 3: Apply the zero-product property to factor and solve the equation.

Let's work through the solution:

Step 1: Observe that each term in the equation 12x39x23x 12x^3 - 9x^2 - 3x has a common factor of 3x 3x . So, we can factor 3x 3x out of the equation, giving us:

3x(4x23x1)=0 3x (4x^2 - 3x - 1) = 0

Step 2: Having factored out 3x 3x , we now have a product of terms equaling zero. According to the zero-product property, at least one of the factors must be zero:

3x=0or4x23x1=0 3x = 0 \quad \text{or} \quad 4x^2 - 3x - 1 = 0

This gives us one solution directly:

x=0 x = 0

Step 3: Solve the quadratic equation 4x23x1=0 4x^2 - 3x - 1 = 0 using the quadratic formula, where a=4 a = 4 , b=3 b = -3 , and c=1 c = -1 :

The quadratic formula is:

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

Applying it to our equation:

x=(3)±(3)244(1)24 x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4 \cdot 4 \cdot (-1)}}}}{2 \cdot 4}

x=3±9+168 x = \frac{{3 \pm \sqrt{{9 + 16}}}}{8}

x=3±258 x = \frac{{3 \pm \sqrt{25}}}{8}

x=3±58 x = \frac{{3 \pm 5}}{8}

This gives us two solutions:

When 25=5 \sqrt{25} = 5 , x=3+58=1 x = \frac{{3 + 5}}{8} = 1 .

When 25=5 \sqrt{25} = -5 , x=358=14 x = \frac{{3 - 5}}{8} = -\frac{1}{4} .

Therefore, the solutions to the equation 12x39x23x=0 12x^3 - 9x^2 - 3x = 0 are x=0 x = 0 , x=1 x = 1 , and x=14 x = -\frac{1}{4} .

Verifying against the provided choices, the correct choice is choice 2: x=0,x=1,x=14 x=0,x=1,x=-\frac{1}{4} .

Answer

x=0,x=1,x=14 x=0,x=1,x=-\frac{1}{4}

Exercise #7

Find the value of the parameter x.

(x4)2+x(x12)=16 (x-4)^2+x(x-12)=16

Video Solution

Step-by-Step Solution

Let's open the parentheses, remembering that there might be more than one solution for the value of X:

(x4)2+x(x12)=16 (x-4)^2+x(x-12)=16

x28x+16+x212x=16 x^2-8x+16+x^2-12x=16

2x220x=0 2x^2-20x=0

2x(x10)=0 2x(x-10)=0

Therefore:

x10=0 x-10=0

x=10 x=10

Or:

2x=0 2x=0

x=0 x=0

Answer

x=0,x=10 x=0,x=10

Exercise #8

Find the value of the parameter x.

2x(3x)+(x3)2=9 -2x(3-x)+(x-3)^2=9

Video Solution

Step-by-Step Solution

To solve the equation 2x(3x)+(x3)2=9-2x(3-x) + (x-3)^2 = 9, follow these steps:

  • Step 1: Expand each term:
    2x(3x)=6x+2x2-2x(3-x) = -6x + 2x^2 and (x3)2=x26x+9(x-3)^2 = x^2 - 6x + 9.
  • Step 2: Substitute the expanded terms into the equation:
    6x+2x2+x26x+9=9-6x + 2x^2 + x^2 - 6x + 9 = 9.
  • Step 3: Combine like terms:
    (2x2+x2)+(6x6x)+9=9(2x^2 + x^2) + (-6x - 6x) + 9 = 9.
  • Step 4: Simplify further:
    3x212x+9=93x^2 - 12x + 9 = 9.
  • Step 5: Move all terms to one side to form a quadratic equation:
    3x212x+99=03x^2 - 12x + 9 - 9 = 0.
  • Step 6: Simplify the expression:
    3x212x=03x^2 - 12x = 0.
  • Step 7: Factor out the common term:
    3x(x4)=03x(x - 4) = 0.
  • Step 8: Solve for x x :
    Since 3x=03x = 0 or x4=0x - 4 = 0, we find x=0x = 0 or x=4x = 4.

Therefore, the values of x x that satisfy the equation are x=0\mathbf{x = 0} and x=4\mathbf{x = 4}.

Answer

x=0,x=4 x=0,x=4