Question

(x²-5x)²+(x²-5x)-12=0​

Answer 1

$(x^2-5x)^2+(x^2-5x)-12=0\\\\t=x^2-5x\; \; ,\; \; \; t^2+t-12=0\; \; ,\; \; t_1=-4\; ,\; t_2=3\; \; (teorema\; Vieta)\\\\a)\; \; x^2-5x=-4\; ,\; \; x^2-5x+4=0\; \; ,\; \; x_1=1\; ,\; x_2=4\\\\b)\; \; x^2-5x=3\; ,\; \; x^2-5x-3=0\; \; ,\; \; x_1=\frac{5-\sqrt{37}}{2}\; ,\; x_2=\frac{5+\sqrt{37}}{2}\\\\Otvet:\; \; x_1=1\; ,\; x_2=4\; ,\; x_3=\frac{5-\sqrt{37}}{2}\; ,\; x_4=\frac{5+\sqrt{37}}{2}\; .$

Answer 2

To solve the quadratic equation $(x^2 - 5x)^2 + (x^2 - 5x) - 12 = 0$, we can use substitution to simplify it and solve for $x$. Let's break it down step by step:

1. Let $y = x^2 - 5x$. This substitution simplifies the equation to $y^2 + y - 12 = 0$.

2. Now we have a standard quadratic equation $y^2 + y - 12 = 0$. We can factor this quadratic equation to find its roots:

   $(y + 4)(y - 3) = 0$

3. Now, we have two possible cases for $y$:

   Case 1: $y + 4 = 0$ => $y = -4$ Case 2: $y - 3 = 0$ => $y = 3$

4. Recall that $y = x^2 - 5x$. So, we can substitute the values of $y$ back into the equation:

   For Case 1: $x^2 - 5x = -4$ For Case 2: $x^2 - 5x = 3$

5. Now, we have two separate quadratic equations:

   For Case 1: $x^2 - 5x + 4 = 0$ For Case 2: $x^2 - 5x - 3 = 0$

6. You can solve each of these quadratic equations using methods such as factoring, completing the square, or using the quadratic formula.

   For Case 1: $x^2 - 5x + 4 = 0$ factors to $(x - 4)(x - 1) = 0$, so the solutions are $x = 4$ and $x = 1$.

   For Case 2: $x^2 - 5x - 3 = 0$ can be solved using the quadratic formula or factoring, and it will have two solutions as well.

Remember that this method gives you the potential solutions for the original equation. Always double-check your solutions by substituting them back into the original equation to make sure they satisfy it.