Question

Помогите решить уравнение |x^2+x-5|=3x

Answer 1

$\mathtt{|x^2+x-5|=3x;~\left[\begin{array}{ccc}\mathtt{x^2+x-5=-3x}\\\mathtt{x^2+x-5=3x}\end{array}\right\left[\begin{array}{ccc}\mathtt{x^2+4x-5=0}\\\mathtt{x^2-2x-5=0}\end{array}\right}\\\\\mathtt{1.~x^2+4x-5=0;~(x+5)(x-1)=0;~x=-5;~1}\\\mathtt{2.~x^2-2x-5=0;~D=(-2)^2-4(-5)=4+20=24=(2\sqrt{6})^2;}\\\mathtt{x=1б\sqrt{6}}$

также не забываем о том, что переменная икс должна быть неотрицательна, так как стоит множителем при тройке, – значении модуля, – который отрицательным быть вовсе не может. Так, нам придётся отсечь 2 отрицательных корня: $\mathtt{-5}$ и $\mathtt{x=1-\sqrt{6}}$

ответ: $\mathtt{x=1;~1-\sqrt{6}}$

Answer 2

Solving the Equation |x^2+x-5|=3x

To solve the equation |x^2+x-5|=3x, we can start by isolating the absolute value expression and then solving for x.

1. Isolating the Absolute Value Expression: We have |x^2+x-5|=3x. To isolate the absolute value expression, we can consider two cases: when the expression inside the absolute value is positive and when it's negative.

2. Case 1: (x^2+x-5) is Positive: When (x^2+x-5) is positive, the equation becomes: x^2+x-5=3x

3. Case 2: (x^2+x-5) is Negative: When (x^2+x-5) is negative, the equation becomes: -(x^2+x-5)=3x

Let's solve each case separately.

Case 1: (x^2+x-5) is Positive

x^2+x-5=3x Rearranging the terms: x^2-2x-5=0

Using the quadratic formula to solve for x: x = (-b ± √(b^2-4ac)) / (2a) where a=1, b=-2, and c=-5

Calculating the discriminant (b^2-4ac): D = (-2)^2 - 4*1*(-5) = 4 + 20 = 24

Since the discriminant is positive, the equation has two real roots: x = (-(-2) ± √24) / (2*1) x = (2 ± 2√6) / 2 x = 1 ± √6

Therefore, the solutions for this case are: x = 1 + √6 and x = 1 - √6

Case 2: (x^2+x-5) is Negative

-(x^2+x-5)=3x Multiplying both sides by -1: x^2+x-5=-3x Rearranging the terms: x^2+4x-5=0

Using the quadratic formula to solve for x: x = (-b ± √(b^2-4ac)) / (2a) where a=1, b=4, and c=-5

Calculating the discriminant (b^2-4ac): D = 4^2 - 4*1*(-5) = 16 + 20 = 36

Since the discriminant is positive, the equation has two real roots: x = (-4 ± √36) / 2 x = (-4 ± 6) / 2 x = 1 and x = -5

Therefore, the solutions for this case are: x = 1 and x = -5

Conclusion

The solutions to the equation |x^2+x-5|=3x are: x = 1 + √6, x = 1 - √6, x = 1, and x = -5.