Question

(х+4)²=3х+40 (2х-3)²=11х-19​

Answer 1

Ответ:

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Answer 2

Ответ:

Пошаговое объяснение:

1)(х+4)²=3х+40

x²+8x+16-3x-40=0

x²+5x-24=0

D=(-5)²-4×1×(-24)=25+96=121

D=121>0   D=√121=11

x=-5±11/2

x₁=-5+11/2=6/2=3

x₂=-5-112=-16/2=-8

2)(2х-3)²=11х-19​

4x²-12x+9-11x+19=0

4x²-23x+28=0

D=(-23)-4×4×28=529-448=81

D=81>0    D=√81=9

x=23±9/8

x₁=23+9/8=32/8=4

x₂=23-9/8=14/8=7/4=1 3/4 или 1.75

Answer 3

To solve the equations, let's expand the square terms and then simplify the expressions to find the value(s) of x that satisfy each equation:

1. (х+4)² = 3х + 40

Expanding the square: (х+4)² = (х+4)(х+4) = х² + 4х + 4х + 16 = х² + 8х + 16

Now, the equation becomes: х² + 8х + 16 = 3х + 40

Moving all terms to one side: х² + 8х - 3х + 16 - 40 = 0

Simplifying: х² + 5х - 24 = 0

Now, we need to solve for x. This is a quadratic equation, and we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

where a = 1, b = 5, and c = -24

x = (-(5) ± √(5² - 4 * 1 * -24)) / 2 * 1 x = (-5 ± √(25 + 96)) / 2 x = (-5 ± √121) / 2 x = (-5 ± 11) / 2

Now, we have two possible solutions:

1. x = (-5 + 11) / 2 x = 6 / 2 x = 3

2. x = (-5 - 11) / 2 x = -16 / 2 x = -8

So, the solutions to the first equation are x = 3 and x = -8.

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2. (2х-3)² = 11х-19

Expanding the square: (2х - 3)² = (2х - 3)(2х - 3) = 4х² - 6х - 6х + 9 = 4х² - 12х + 9

Now, the equation becomes: 4х² - 12х + 9 = 11х - 19

Moving all terms to one side: 4х² - 12х - 11х + 9 + 19 = 0

Simplifying: 4х² - 23х + 28 = 0

We again have a quadratic equation. Let's use the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

where a = 4, b = -23, and c = 28

x = (-(−23) ± √((−23)² - 4 * 4 * 28)) / 2 * 4 x = (23 ± √(529 - 448)) / 8 x = (23 ± √81) / 8 x = (23 ± 9) / 8

Now, we have two possible solutions:

1. x = (23 + 9) / 8 x = 32 / 8 x = 4

2. x = (23 - 9) / 8 x = 14 / 8 x = 7/4

So, the solutions to the second equation are x = 4 and x = 7/4.