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Zorluk: OrtaSolving Quadratic Equations by Factoring

What is the positive difference between the two real solutions to the equation (x1)2=5x5(x - 1)^2 = 5x - 5?

  1. A
    1
  2. 5Cevap
  3. C
    7
  4. D
    6
  5. E
    4

Cevap

The positive difference between the two real solutions is 5.
Expanding the left side of (x1)2=5x5(x - 1)^2 = 5x - 5 yields x22x+1=5x5x^2 - 2x + 1 = 5x - 5. Moving all terms to the left side by subtracting 5x5x and adding 55 gives the standard quadratic equation x27x+6=0x^2 - 7x + 6 = 0. Factoring this expression gives (x1)(x6)=0(x - 1)(x - 6) = 0, which yields the solutions 11 and 66. The positive difference between these solutions is 61=56 - 1 = 5.

Adım Adım Çözüm

1
Expand the squared binomial on the left side of the equation.
x22x+1=5x5x^2 - 2x + 1 = 5x - 5
Before factoring a quadratic equation, all terms must be expanded and moved to one side to set the equation equal to zero.
2
Subtract 5x5x and add 55 to both sides to rearrange the equation into standard quadratic form.
x27x+6=0x^2 - 7x + 6 = 0
This sets the equation equal to zero, which is a prerequisite for using the zero product property.
3
Factor the quadratic expression by finding two numbers that multiply to 66 and add to 7-7.
(x1)(x6)=0(x - 1)(x - 6) = 0
Factoring allows us to split the quadratic equation into two linear equations.
4
Set each factor to zero to solve for xx.
x=1x = 1 and x=6x = 6
By the zero product property, if the product of two factors is zero, at least one of the factors must be zero.
5
Subtract the smaller solution from the larger solution to find the positive difference.
61=56 - 1 = 5
The question asks for the positive difference between the two solutions.

Anahtar Kavram

Solving quadratic equations by expanding, rearranging into standard form, and factoring over the integers.
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