1. **Problem statement:** Find the values of $k$ for which each quadratic equation has no real roots. 2. **Concept:** A quadratic equation $ax^2 + bx + c = 0$ has no real roots if its discriminant is less than zero. The discriminant formula is: $$\Delta = b^2 - 4ac$$ If $\Delta < 0$, the quadratic has no real roots. --- ### a) $kx^2 - 4x + 8 = 0$ - Here, $a = k$, $b = -4$, $c = 8$ - Discriminant: $$\Delta = (-4)^2 - 4(k)(8) = 16 - 32k$$ - For no real roots: $$16 - 32k < 0$$ $$16 < 32k$$ $$\frac{16}{32} < k$$ $$\frac{1}{2} < k$$ --- ### b) $3x^2 + 5x + k + 1 = 0$ - Here, $a = 3$, $b = 5$, $c = k + 1$ - Discriminant: $$\Delta = 5^2 - 4(3)(k+1) = 25 - 12(k+1) = 25 - 12k - 12 = 13 - 12k$$ - For no real roots: $$13 - 12k < 0$$ $$13 < 12k$$ $$\frac{13}{12} < k$$ --- ### c) $2x^2 + 8x - 5 = kx^2$ - Rearrange to standard form: $$2x^2 + 8x - 5 - kx^2 = 0$$ $$ (2 - k)x^2 + 8x - 5 = 0$$ - Here, $a = 2 - k$, $b = 8$, $c = -5$ - Discriminant: $$\Delta = 8^2 - 4(2-k)(-5) = 64 + 20(2-k) = 64 + 40 - 20k = 104 - 20k$$ - For no real roots: $$104 - 20k < 0$$ $$104 < 20k$$ $$\frac{104}{20} < k$$ $$5.2 < k$$ --- ### d) $2x^2 + k = 3(x - 2)$ - Expand right side: $$2x^2 + k = 3x - 6$$ - Rearrange: $$2x^2 - 3x + k + 6 = 0$$ - Here, $a = 2$, $b = -3$, $c = k + 6$ - Discriminant: $$\Delta = (-3)^2 - 4(2)(k+6) = 9 - 8(k+6) = 9 - 8k - 48 = -8k - 39$$ - For no real roots: $$-8k - 39 < 0$$ $$-8k < 39$$ $$k > -\frac{39}{8}$$ --- ### e) $kx^2 + 2kx = 4x - 6$ - Rearrange: $$kx^2 + 2kx - 4x + 6 = 0$$ $$kx^2 + (2k - 4)x + 6 = 0$$ - Here, $a = k$, $b = 2k - 4$, $c = 6$ - Discriminant: $$\Delta = (2k - 4)^2 - 4(k)(6) = (2k - 4)^2 - 24k$$ - Expand: $$ (2k - 4)^2 = 4k^2 - 16k + 16$$ - So: $$\Delta = 4k^2 - 16k + 16 - 24k = 4k^2 - 40k + 16$$ - For no real roots: $$4k^2 - 40k + 16 < 0$$ - Divide entire inequality by 4: $$k^2 - 10k + 4 < 0$$ - Find roots of quadratic $k^2 - 10k + 4 = 0$: $$k = \frac{10 \pm \sqrt{100 - 16}}{2} = \frac{10 \pm \sqrt{84}}{2} = \frac{10 \pm 2\sqrt{21}}{2} = 5 \pm \sqrt{21}$$ - Since $k^2 - 10k + 4$ is a positive leading coefficient quadratic, it is less than zero between its roots: $$5 - \sqrt{21} < k < 5 + \sqrt{21}$$ --- ### f) $kx^2 + kx = 3x - 2$ - Rearrange: $$kx^2 + kx - 3x + 2 = 0$$ $$kx^2 + (k - 3)x + 2 = 0$$ - Here, $a = k$, $b = k - 3$, $c = 2$ - Discriminant: $$\Delta = (k - 3)^2 - 4(k)(2) = (k - 3)^2 - 8k$$ - Expand: $$ (k - 3)^2 = k^2 - 6k + 9$$ - So: $$\Delta = k^2 - 6k + 9 - 8k = k^2 - 14k + 9$$ - For no real roots: $$k^2 - 14k + 9 < 0$$ - Find roots of quadratic $k^2 - 14k + 9 = 0$: $$k = \frac{14 \pm \sqrt{196 - 36}}{2} = \frac{14 \pm \sqrt{160}}{2} = \frac{14 \pm 4\sqrt{10}}{2} = 7 \pm 2\sqrt{10}$$ - Since leading coefficient is positive, inequality holds between roots: $$7 - 2\sqrt{10} < k < 7 + 2\sqrt{10}$$ --- **Final answers:** - a) $k > \frac{1}{2}$ - b) $k > \frac{13}{12}$ - c) $k > 5.2$ - d) $k > -\frac{39}{8}$ - e) $5 - \sqrt{21} < k < 5 + \sqrt{21}$ - f) $7 - 2\sqrt{10} < k < 7 + 2\sqrt{10}$