Laswerk Vir watter waarde (s) van $$ k $$ Sal $$ 2 x^{2}-2 x-k=0 $$ gelyk wortels het 2. Bereken die waarde van $$ k $$, as $$ 3 x^{2}-k x+3=0 $$ rëel wortels het. 3. Vir watter waarde sal $$ 4 x^{2}+3 x+p=0 $$ : a) gelyk wortels het? b) ongelyk wortels hat? 4. Vir watter waarde van $$ m $$ sal $$ (m-2) x^{2}-x-m x=7-2 m $$

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Alright, I need to solve these quadratic equation problems step by step. Let's tackle them one by one. **Problem 1:** Find the value(s) of $$ k $$ such that the equation $$ 2x^{2} - 2x - k = 0 $$ has equal roots. To find the value of $$ k $$ for which the quadratic equation has equal roots, I'll use the discriminant formula. The discriminant $$ D $$ of a quadratic equation $$ ax^{2} + bx + c = 0 $$ is given by: $$ D = b^{2} - 4ac $$ For the equation to have equal roots, the discriminant must be zero: $$ D = 0 $$ Given the equation $$ 2x^{2} - 2x - k = 0 $$, let's identify the coefficients: $$ a = 2,\quad b = -2,\quad c = -k $$ Plugging these into the discriminant formula: $$ D = (-2)^{2} - 4(2)(-k) = 4 + 8k = 0 $$ Solving for $$ k $$: $$ 8k = -4 \ k = -\frac{4}{8} = -\frac{1}{2} $$ So, $$ k = -\frac{1}{2} $$ is the value that makes the equation have equal roots. **Problem 2:** Find the value of $$ k $$ such that the equation $$ 3x^{2} - kx + 3 = 0 $$ has real roots. For a quadratic equation to have real roots, the discriminant must be greater than or equal to zero: $$ D \geq 0 $$ Using the discriminant formula: $$ D = (-k)^{2} - 4(3)(3) = k^{2} - 36 \geq 0 $$ Solving the inequality: $$ k^{2} \geq 36 \ k \leq -6 \quad \text{or} \quad k \geq 6 $$ Therefore, $$ k $$ must be less than or equal to -6 or greater than or equal to 6 for the equation to have real roots. **Problem 3:** For the equation $$ 4x^{2} + 3x + p = 0 $$: a) Find the value(s) of $$ p $$ for which the equation has equal roots. b) Find the value(s) of $$ p $$ for which the equation has unequal roots. **Part a:** For equal roots, the discriminant must be zero: $$ D = 3^{2} - 4(4)(p) = 9 - 16p = 0 \ 16p = 9 \ p = \frac{9}{16} $$ So, $$ p = \frac{9}{16} $$ ensures equal roots. **Part b:** For unequal roots, the discriminant must be greater than zero: $$ D > 0 \ 9 - 16p > 0 \ 16p < 9 \ p < \frac{9}{16} $$ Therefore, $$ p $$ must be less than $$ \frac{9}{16} $$ for the equation to have unequal roots. **Problem 4:** Find the value(s) of $$ m $$ such that the equation $$ (m - 2)x^{2} - x - m = 7 - 2m $$ holds true. First, let's simplify the equation: $$ (m - 2)x^{2} - x - m = 7 - 2m \ (m - 2)x^{2} - x - m + 2m = 7 \ (m - 2)x^{2} - x + m = 7 $$ To find the value(s) of $$ m $$, we need to set the equation to zero: $$ (m - 2)x^{2} - x + m - 7 = 0 $$ This is a quadratic in terms of $$ x $$, but since $$ m $$ is a variable, we need to find $$ m $$ such that the equation holds for all $$ x $$. For this to be true, the coefficients of like terms must satisfy certain conditions. Let's equate the coefficients of $$ x^{2} $$, $$ x $$, and the constant term to zero: $$ m - 2 = 0 \quad \Rightarrow \quad m = 2 \ -1 = 0 \quad \Rightarrow \quad \text{This is impossible} \ m - 7 = 0 \quad \Rightarrow \quad m = 7 $$ There's a contradiction here because $$ m $$ cannot simultaneously be 2 and 7. Therefore, there is no real value of $$ m $$ that satisfies the equation for all $$ x $$. Alternatively, if we interpret the problem as finding $$ m $$ such that the equation has real roots for specific $$ x $$, we can use the discriminant: $$ D = (-1)^{2} - 4(m - 2)(m - 7) = 1 - 4(m^{2} - 9m + 14) = 1 - 4m^{2} + 36m - 56 = -4m^{2} + 36m - 55 $$ For real roots: $$ -4m^{2} + 36m - 55 \geq 0 \ 4m^{2} - 36m + 55 \leq 0 $$ Solving the quadratic inequality: $$ m = \frac{36 \pm \sqrt{(-36)^{2} - 4(4)(55)}}{2(4)} = \frac{36 \pm \sqrt{1296 - 880}}{8} = \frac{36 \pm \sqrt{416}}{8} = \frac{36 \pm 4\sqrt{26}}{8} = \frac{9 \pm \sqrt{26}}{2} $$ So, $$ m $$ must be between $$ \frac{9 - \sqrt{26}}{2} $$ and $$ \frac{9 + \sqrt{26}}{2} $$ for the equation to have real roots. 1. $$ k = -\frac{1}{2} $$ 2. $$ k \leq -6 $$ or $$ k \geq 6 $$ 3. a) $$ p = \frac{9}{16} $$ b) $$ p < \frac{9}{16} $$ 4. There is no real value of $$ m $$ that satisfies the equation for all $$ x $$. For real roots, $$ m $$ must be between $$ \frac{9 - \sqrt{26}}{2} $$ and $$ \frac{9 + \sqrt{26}}{2} $$.

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