This worksheet contains quadratic equations, where you are tasked with solving using the discriminant to determine the number of real roots, and applying the quadratic formula to find the solutions. I'll help break this down into parts:

Step 1: Use the Discriminant (Δ = b² - 4ac)

The discriminant helps determine:

• Δ > 0: Two distinct real roots.
• Δ = 0: One real root (repeated root).
• Δ < 0: No real roots (complex roots).

Let’s calculate the discriminant for each missing row:

1. $5x^2 - 2x + 25 = 0$

   • $a = 5$, $b = -2$, $c = 25$
   • $\Delta = b^2 - 4ac = (-2)^2 - 4(5)(25) = 4 - 500 = -496$
   • Number of Roots: 0 (No real roots)
   • Justification: The discriminant is negative, so the graph does not intersect the x-axis.

2. $4x^2 + 11x + 1 = 0$

   • $a = 4$, $b = 11$, $c = 1$
   • $\Delta = b^2 - 4ac = (11)^2 - 4(4)(1) = 121 - 16 = 105$
   • Number of Roots: 2 (Two distinct real roots)
   • Justification: The discriminant is positive, so the graph intersects the x-axis twice.

Step 2: Solve Using the Quadratic Formula

The quadratic formula is: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$ Now let's solve each of the quadratic equations in the second section:

1. $2x^2 - 11x + 10 = 0$

   • $a = 2, b = -11, c = 10$
   • $\Delta = (-11)^2 - 4(2)(10) = 121 - 80 = 41$
   • Roots: $x = \frac{-(-11) \pm \sqrt{41}}{2(2)} = \frac{11 \pm \sqrt{41}}{4}$ Approximate roots: $x \approx \frac{11 + 6.4}{4} = 4.35 \quad \text{and} \quad x \approx \frac{11 - 6.4}{4} = 1.15$

2. $x^2 + 14x + 49 = 0$

   • $a = 1, b = 14, c = 49$
   • $\Delta = (14)^2 - 4(1)(49) = 196 - 196 = 0$
   • Roots: $x = \frac{-14 \pm \sqrt{0}}{2(1)} = \frac{-14}{2} = -7$ Single root: $x = -7$

3. $x^2 + 8x + 19 = 0$

   • $a = 1, b = 8, c = 19$
   • $\Delta = (8)^2 - 4(1)(19) = 64 - 76 = -12$
   • No real roots, since $\Delta < 0$.

4. $x^2 - 6x + 9 = 0$

   • $a = 1, b = -6, c = 9$
   • $\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0$
   • Roots: $x = \frac{-(-6) \pm \sqrt{0}}{2(1)} = \frac{6}{2} = 3$ Single root: $x = 3$

5. $2x^2 - 7x = 16$

   • Rearrange: $2x^2 - 7x - 16 = 0$
   • $a = 2, b = -7, c = -16$
   • $\Delta = (-7)^2 - 4(2)(-16) = 49 + 128 = 177$
   • Roots: $x = \frac{-(-7) \pm \sqrt{177}}{2(2)} = \frac{7 \pm \sqrt{177}}{4}$ Approximate roots: $x \approx \frac{7 + 13.3}{4} = 5.08 \quad \text{and} \quad x \approx \frac{7 - 13.3}{4} = -1.58$

6. $3x^2 = -4(x + 3)$

   • Expand: $3x^2 + 4x + 12 = 0$
   • $a = 3, b = 4, c = 12$
   • $\Delta = (4)^2 - 4(3)(12) = 16 - 144 = -128$
   • No real roots, since $\Delta < 0$.

Would you like detailed steps for a specific calculation?

5 Relative Questions

1. How does the discriminant affect the nature of roots geometrically?
2. Can we always use the quadratic formula regardless of the discriminant's value?
3. How do repeated roots (Δ = 0) relate to the vertex of the parabola?
4. Why does a negative discriminant imply no real solutions?
5. What alternative methods can solve quadratic equations besides the quadratic formula?

Tip:

Always simplify equations to standard form ($ax^2 + bx + c = 0$) before calculating the discriminant or using the quadratic formula!