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22 votes
22 votes
I need to know how to solve them with work shown

I need to know how to solve them with work shown-example-1
I need to know how to solve them with work shown-example-1
I need to know how to solve them with work shown-example-2
I need to know how to solve them with work shown-example-3
User Eliatou
by
2.9k points

2 Answers

17 votes
17 votes

#1

  • (x-2)²=25
  • (x-2)²=5²
  • x-2=5
  • x=5+2
  • x=7

#2

  • (x-4)²-8=8
  • (x-4)²=8+8=16
  • (x-4)²=4²
  • x-4=4
  • x=4+4
  • x=8
User Roger Dwan
by
2.4k points
8 votes
8 votes

Explanation:

For the first one,

(x-2)² = 25

or, x² - 4x + 4 = 25

or, x² - 4x + 4 - 25 =0

or, x² - 4x - 21 = 0

or, x² - (7 - 3)x - 21 = 0

or, x² - 7x + 3x - 21 = 0

or, x(x - 7) + 3(x - 7) = 0

or, (x - 7) (x + 3) = 0

For RHS to be zero, either one of the factor should be zero,

So either,

x - 7 = 0

so, x = 7

Or,

x + 3 = 0

so, x = -3

x have 2 values because both of these values satisfies the given equation.

For second one,

(x - 4)² - 8 = 8

or, (x - 4)² = 8+8

or, x² - 8x + 16 = 16

or, x² - 8x + 16 - 16 = 0

or, x² - 8x = 0

or, x(x- 8) = 0

For RHS to be zero, either x should be zero or (x - 8) should be zero,

So either,

x = 0

or,

x - 8 = 0

so, x = 8

x have two values because both of the x's values satisfies the equation

For last one :-

x - 6x = -7

or, x - 6x + 7 = 0

so,

a = 1, b = -6 and c = 7

Use quadratic equation and substitute the values in given formula you will get the answer.

(I can't type the equation here so I can't solve it step by step)

User Naz
by
2.6k points