2. For which values of \( k \) will the following equations have cgual rools? \( 2.1 \quad 2 x^{2}+3 x-2=k \) 3. Determine the value(s) of \( k \) if the roots of \( x^{2}-2 x=4-k \) are real. 4. Prove that the roots of \( (a-1) x^{2}+2 a x-x+2=0 \) are real and rational.

**Problem 2** We start with the equation \[ 2x^2 + 3x - 2 = k. \] Rearrange it into the standard form of a quadratic equation by subtracting \( k \) from both sides: \[ 2x^2 + 3x - (2+k) = 0. \] For a quadratic equation of the form \[ ax^2 + bx + c = 0, \] the condition for equal roots is that the discriminant is zero. The discriminant is given by \[ \Delta = b^2 - 4ac. \] Here, \[ a = 2, \quad b = 3, \quad c = -2-k. \] Thus, we require: \[ \Delta = 3^2 - 4(2)(-2-k) = 0. \] Calculate step by step: \[ 9 - 4(2)(-2-k) = 9 + 8(2+k) = 9 + 16 + 8k = 25 + 8k. \] Setting it equal to zero: \[ 25 + 8k = 0 \quad \Longrightarrow \quad 8k = -25 \quad \Longrightarrow \quad k = -\frac{25}{8}. \] --- **Problem 3** The given equation is \[ x^2 - 2x = 4 - k. \] Bring all terms to one side to write it in standard form: \[ x^2 -2x - (4-k) = 0 \quad \Longrightarrow \quad x^2 - 2x - 4 + k = 0. \] Here, the coefficients are: \[ a = 1,\quad b = -2,\quad c = k-4. \] For the roots to be real, the discriminant must be non-negative: \[ \Delta = b^2 - 4ac \geq 0. \] Substitute the coefficients: \[ \Delta = (-2)^2 - 4(1)(k-4) = 4 - 4(k-4). \] Simplify the expression: \[ 4 - 4(k-4) = 4 - 4k + 16 = 20 - 4k. \] Set the inequality: \[ 20 - 4k \geq 0 \quad \Longrightarrow \quad 4k \leq 20 \quad \Longrightarrow \quad k \leq 5. \] --- **Problem 4** The given equation is \[ (a-1)x^2 + 2ax - x + 2 = 0. \] First, combine like terms in the \( x \)-linear coefficient: \[ 2ax - x = (2a-1)x. \] Thus, the equation can be rewritten as: \[ (a-1)x^2 + (2a-1)x + 2 = 0. \] The coefficients are: \[ a_1 = a-1, \quad b_1 = 2a-1, \quad c_1 = 2. \] The discriminant is: \[ \Delta = b_1^2 - 4a_1c_1 = (2a-1)^2 - 4(a-1)(2). \] Expand and simplify the expression: \[ (2a-1)^2 = 4a^2 - 4a + 1, \] and \[ 4(a-1)(2)=8(a-1)=8a-8. \] Thus: \[ \Delta = 4a^2 - 4a + 1 - 8a + 8 = 4a^2 - 12a + 9. \] Notice that: \[ 4a^2 - 12a + 9 = (2a-3)^2. \] Since \((2a-3)^2\) is a perfect square, it is always non-negative regardless of the value of \( a \). Therefore, the roots of the equation are real. Furthermore, because the discriminant is a perfect square, the roots obtained from the quadratic formula: \[ x = \frac{-(2a-1) \pm |2a-3|}{2(a-1)} \] are rational provided that \( a \) is a rational number (and \( a \neq 1 \) to avoid a degenerate case). This demonstrates that the roots are not only real but also rational.