Question

1) x^2-5|x|=0 2) 2x^2+|x|-3x=0 3) x^2+|x+4|=4 4) x^2-|x-5|=5 Сделайте пожалуйста

Answer 1

$1) \;\; x^2-5\mid x \mid=0 \\ \\ x^2-5x=0=\ \textgreater \ x(x-5)=0=\ \textgreater \ x=0 \;\; x-5=0=\ \textgreater \ x=0 \;\; x=5 \\ x^2-5*(-x)=0=\ \textgreater \ x(x+5)=0=\ \textgreater \ x=0 \;\; x+5=0=\ \textgreater \ x=-5 \\ \\ \boxed{x=0 \;\; x=5 \;\; x=-5} \\ \\ \\ 2) \;\; 2x^2+\mid x \mid -3x=0 \\ \\ 2x^2+x-3x=0=\ \textgreater \ 2x^2-2x=0 =\ \textgreater \ 2x(x-1)=0=\ \textgreater \ x=0 \;x=1\\ 2x^2-x-3x=0=\ \textgreater \ 2x^2-4x=0=\ \textgreater \ 2x(x-2)=0-\ \textgreater \ x=0 \; x=2 \\ \\ No \;\; tak \;\; kak \;\; vo \;\; vtorom \;\; x\ \textless \ 0 \;\; znachit \;\; x\notin R \\ \\ \boxed {x=0 \;\; x=1} \\ \\ \\ 3)\;\; x^2+\mid x+4 \mid =4 \\ \\ x^2+x+4=4,x \geq -4 \\ x^2-(x+4)=4, x+4 \leq -4 \\ \\ x^2+x=0=\ \textgreater \ x(x+1)=0=\ \textgreater \ x=0 \;\; x=-1 \\ \\ vo \;\; vtorom \;\; ne \;\; budet \;\; kornei\;\; tak \;\; kak \;\; x \leq -4 \\ \\ \boxed {x=-1 \;\; x=0} \\ \\ \\ 4)\;\; x^2-\mid x-5 \mid =5 \\ \\ x^2-(x-5)=5 ,x \geq 5 \\ x^2-(-(x-5))=5,x\ \textless \ 5 \\ \\ x^2-x+5=5=\ \textgreater \ x^2-x=0=\ \textgreater \ x=0 \;\; x=1 \;\; NO \;\; x \geq 5 \\ \\ x^2+(x-5)=5=\ \textgreater \ x^2+x-5=5=\ \textgreater \ x^2+x-10=0 \\ \\ D=41; \sqrt{D}=\sqrt{41} \\ \\ \boxed {x= \frac{1\pm \sqrt{41} }{2}}$

Answer 2

Equation 1: x^2 - 5|x| = 0

To solve this equation, we need to consider two cases: when x is positive and when x is negative.

Case 1: x is positive When x is positive, the equation becomes: x^2 - 5x = 0

Factoring out x, we get: x(x - 5) = 0

So, either x = 0 or x - 5 = 0. Solving for x, we find two solutions: x = 0 or x = 5.

Case 2: x is negative When x is negative, the equation becomes: x^2 + 5x = 0

Factoring out x, we get: x(x + 5) = 0

So, either x = 0 or x + 5 = 0. Solving for x, we find two solutions: x = 0 or x = -5.

Therefore, the solutions to the equation x^2 - 5|x| = 0 are x = 0, x = 5, and x = -5.

Equation 2: 2x^2 + |x| - 3x = 0

To solve this equation, we need to consider two cases: when x is positive and when x is negative.

Case 1: x is positive When x is positive, the equation becomes: 2x^2 + x - 3x = 0

Combining like terms, we get: 2x^2 - 2x = 0

Factoring out 2x, we get: 2x(x - 1) = 0

So, either 2x = 0 or x - 1 = 0. Solving for x, we find two solutions: x = 0 or x = 1.

Case 2: x is negative When x is negative, the equation becomes: 2x^2 - x - 3x = 0

Combining like terms, we get: 2x^2 - 4x = 0

Factoring out 2x, we get: 2x(x - 2) = 0

So, either 2x = 0 or x - 2 = 0. Solving for x, we find two solutions: x = 0 or x = 2.

Therefore, the solutions to the equation 2x^2 + |x| - 3x = 0 are x = 0, x = 1, and x = 2.

Equation 3: x^2 + |x + 4| = 4

To solve this equation, we need to consider two cases: when x + 4 is positive and when x + 4 is negative.

Case 1: x + 4 is positive When x + 4 is positive, the equation becomes: x^2 + (x + 4) = 4

Expanding the equation, we get: x^2 + x + 4 = 4

Simplifying, we have: x^2 + x = 0

Factoring out x, we get: x(x + 1) = 0

So, either x = 0 or x + 1 = 0. Solving for x, we find two solutions: x = 0 or x = -1.

Case 2: x + 4 is negative When x + 4 is negative, the equation becomes: x^2 - (x + 4) = 4

Expanding the equation, we get: x^2 - x - 4 = 4

Simplifying, we have: x^2 - x - 8 = 0

Using the quadratic formula, we find two solutions: x ≈ -2.828 or x ≈ 2.828.

Therefore, the solutions to the equation x^2 + |x + 4| = 4 are x = 0, x = -1, x ≈ -2.828, and x ≈ 2.828.

Equation 4: x^2 - |x - 5| = 5

To solve this equation, we need to consider two cases: when x - 5 is positive and when x - 5 is negative.

Case 1: x - 5 is positive When x - 5 is positive, the equation becomes: x^2 - (x - 5) = 5

Expanding the equation, we get: x^2 - x + 5 = 5

Simplifying, we have: x^2 - x = 0

Factoring out x, we get: x(x - 1) = 0

So, either x = 0 or x - 1 = 0. Solving for x, we find two solutions: x = 0 or x = 1.

Case 2: x - 5 is negative When x - 5 is negative, the equation becomes: x^2 + (x - 5) = 5

Expanding the equation, we get: x^2 + x - 5 = 5

Simplifying, we have: x^2 + x - 10 = 0

Using the quadratic formula, we find two solutions: x ≈ -3.162 or x ≈ 2.162.

Therefore, the solutions to the equation x^2 - |x - 5| = 5 are x = 0, x = 1, x ≈ -3.162, and x ≈ 2.162.