What are the possible values of x if (4x – 5)2 = 49? Check all that apply. -4/5
-1/2
3
05
07

Hey there,

What are the possible values of x?

(4x - 5)² = 49

→ (4x)² - 2*4x*5 + 5² = 49

→ 16x² - 40x + 25 = 49

→ 16x² - 40x + 25 - 49 = 0

→ 16x² - 40x - 24 = 0

∆ = b² - 4ac

∆ = (-40)² - 4*16*(-24)

∆ = 1600 - 64*(-24)

∆ = 1600 - (-1536)

∆ = 1600 + 1536

∆ = 3136

∆ = 3136 > 0; There are two real solutions:

x1 = (-b - √∆)/2a = (40-56)/32 = -16/32 = -1/2

x2 = (-b + √∆)/2a = (40+56)/32 = 96/32 = 3

S={ -1/2 ; 3 }

✅✅;)

Hey there!

(4x - 5)^2 = 49

(4x - 5)(4x - 5) = 49

16x^2 - 40x + 25 = 49

SUBTRACT 49 to BOTH SIDES

16x^2 - 40x + 25 - 49 = 49 - 49

SIMPLIFY IT!

16x^2 - 40x - 24 = 0

[FACTOR the LEFT SIDE] of THE EQUATION

We get: 8(2x + 1)(x - 3) = 0

SET EACH FACTOR to EQUAL 0

We get: 2x + 1 = 0 OR x - 3 = 0

SIMPLIFY IT!

x = - 1/2

≈ x = -0.5

OR

x = 3

Therefore, your answer is: x = -1/2 & x = 3

(Choice B. & Choice C.)

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)