1. **Problem statement:** Find the values of $k$ for which each quadratic equation has two distinct roots. 2. **Key concept:** A quadratic equation $ax^2 + bx + c = 0$ has two distinct roots if its discriminant $\Delta = b^2 - 4ac > 0$. --- ### a) $x^2 + 8x + 3 = k$ Rewrite as $x^2 + 8x + (3 - k) = 0$. - Here, $a=1$, $b=8$, $c=3-k$. - Discriminant: $$\Delta = 8^2 - 4 \times 1 \times (3-k) = 64 - 4(3-k) = 64 - 12 + 4k = 52 + 4k$$ - For two distinct roots: $$52 + 4k > 0$$ - Solve: $$4k > -52 \Rightarrow k > -13$$ --- ### b) $2x^2 - 5x = 4 - k$ Rewrite as $2x^2 - 5x - (4 - k) = 0$ or $2x^2 - 5x - 4 + k = 0$. - Here, $a=2$, $b=-5$, $c=-4 + k$. - Discriminant: $$\Delta = (-5)^2 - 4 \times 2 \times (-4 + k) = 25 - 8(-4 + k) = 25 + 32 - 8k = 57 - 8k$$ - For two distinct roots: $$57 - 8k > 0$$ - Solve: $$-8k > -57 \Rightarrow k < \frac{57}{8} = 7.125$$ --- ### c) $kx^2 - 4x + 2 = 0$ - Here, $a=k$, $b=-4$, $c=2$. - Discriminant: $$\Delta = (-4)^2 - 4 \times k \times 2 = 16 - 8k$$ - For two distinct roots: $$16 - 8k > 0$$ - Solve: $$-8k > -16 \Rightarrow k < 2$$ - Also, $a=k \neq 0$ for it to be quadratic, so $k \neq 0$. --- ### d) $kx^2 + 2(k - 1)x + k = 0$ - Here, $a=k$, $b=2(k-1)$, $c=k$. - Discriminant: $$\Delta = [2(k-1)]^2 - 4 \times k \times k = 4(k-1)^2 - 4k^2 = 4[(k-1)^2 - k^2]$$ - Expand inside bracket: $$(k-1)^2 - k^2 = (k^2 - 2k + 1) - k^2 = -2k + 1$$ - So, $$\Delta = 4(-2k + 1) = 4 - 8k$$ - For two distinct roots: $$4 - 8k > 0$$ - Solve: $$-8k > -4 \Rightarrow k < \frac{1}{2}$$ - Also, $k \neq 0$ for quadratic. --- ### e) $2x^2 = 2(x - 1) + k$ Rewrite as $$2x^2 - 2x + 2 - k = 0$$ - Here, $a=2$, $b=-2$, $c=2 - k$. - Discriminant: $$\Delta = (-2)^2 - 4 \times 2 \times (2 - k) = 4 - 8(2 - k) = 4 - 16 + 8k = -12 + 8k$$ - For two distinct roots: $$-12 + 8k > 0$$ - Solve: $$8k > 12 \Rightarrow k > \frac{12}{8} = 1.5$$ --- ### f) $kx^2 + (2k - 5)x = 1 - k$ Rewrite as $$kx^2 + (2k - 5)x - (1 - k) = 0$$ - Here, $a=k$, $b=2k - 5$, $c= -1 + k$. - Discriminant: $$\Delta = (2k - 5)^2 - 4 \times k \times (-1 + k) = (2k - 5)^2 + 4k(1 - k)$$ - Expand: $$(2k - 5)^2 = 4k^2 - 20k + 25$$ - So, $$\Delta = 4k^2 - 20k + 25 + 4k - 4k^2 = -16k + 25$$ - For two distinct roots: $$-16k + 25 > 0$$ - Solve: $$-16k > -25 \Rightarrow k < \frac{25}{16} = 1.5625$$ - Also, $k \neq 0$ for quadratic. --- **Final answers:** - a) $k > -13$ - b) $k < 7.125$ - c) $k < 2$ and $k \neq 0$ - d) $k < 0.5$ and $k \neq 0$ - e) $k > 1.5$ - f) $k < 1.5625$ and $k \neq 0$