Solution of quadratic equation using graph

Question 1: Projectile Motion of a Stone (from file image_00117f.png)

The Question: A stone is thrown with an initial speed $u$ (m/s). The height of the stone’s trajectory above the ground is, $H(t) = -5t^2 + \frac{1}{2}ut$ (here $t$ is the time of flight). If the highest point in air that the stone can reach is 5m above the ground, then calculate the initial speed $u$.

Core Concept: Maximum Value of a Quadratic Function

The path of a projectile under gravity is a parabola. The function given, $H(t) = -5t^2 + \frac{1}{2}ut$, is a quadratic equation where the coefficient of the $t^2$ term is negative (-5). This means the parabola opens downwards ($\cap$), and its vertex represents the highest point of the trajectory.

We can find the time ($t$) it takes to reach this vertex and the maximum height (the y-value of the vertex) using the vertex formula. For a function $y = at^2 + bt + c$:

• Time to reach vertex: $t_{vertex} = -\frac{b}{2a}$
• The maximum height is the value of the function at that time, $H(t_{vertex})$.

Detailed Solution:

1. Identify the coefficients from the function $H(t) = -5t^2 + (\frac{u}{2})t$:

   • $a = -5$
   • $b = \frac{u}{2}$

2. Find the time ($t$) to reach the highest point:

   • $t_{vertex} = -\frac{b}{2a} = -\frac{u/2}{2(-5)} = -\frac{u/2}{-10} = \frac{u}{20}$

3. Use the given maximum height to find $u$:

   • We are told the maximum height is 5m. This means when $t = \frac{u}{20}$, the value of $H(t)$ is 5.
   • Let’s substitute $t = \frac{u}{20}$ back into the original equation and set it equal to 5: $$5 = -5\left(\frac{u}{20}\right)^2 + \frac{u}{2}\left(\frac{u}{20}\right)$$

4. Solve the equation for $u$:

   $$5 = -5\left(\frac{u^2}{400}\right) + \frac{u^2}{40}$$
   $$5 = -\frac{5u^2}{400} + \frac{u^2}{40}$$
   • Simplify the first fraction: $$5 = -\frac{u^2}{80} + \frac{u^2}{40}$$
   • Find a common denominator (80): $$5 = -\frac{u^2}{80} + \frac{2u^2}{80}$$$$5 = \frac{-u^2 + 2u^2}{80}$$$$5 = \frac{u^2}{80}$$
   • Isolate $u^2$: $$u^2 = 5 \times 80 = 400$$
   • Take the square root (speed must be positive): $$u = \sqrt{400} = 20$$

Final Answer: The initial speed $u$ is 20 m/s.

Question 2: Finding Roots of a Quadratic Equation (from file image_000ddb.png)

The Question: Find the roots of the equation: $x^2 - 8x + 15 = 0$.

Core Concept: Roots of an Equation

The “roots” (or “solutions”) of an equation are the values of the variable (in this case, $x$) that make the statement true. For a quadratic equation, we can find the roots by factoring or by using the quadratic formula. Factoring is often quicker if the numbers are simple.

Method: Factoring To factor an equation of the form $x^2 + bx + c = 0$, we look for two numbers that multiply to give $c$ and add to give $b$.

Detailed Solution:

1. Analyze the equation: $x^2 - 8x + 15 = 0$.
   • We need two numbers that multiply to +15.
   • We need the same two numbers to add to -8.
2. Find the two numbers:
   • Factors of 15: (1, 15), (3, 5).
   • To get a positive product and a negative sum, both numbers must be negative.
   • Let’s try (-3) and (-5):
     • $(-3) \times (-5) = +15$ (This works)
     • $(-3) + (-5) = -8$ (This also works)
3. Rewrite the equation in factored form:
   • The equation becomes $(x - 3)(x - 5) = 0$.
4. Solve for the roots:
   • For the product of two terms to be zero, at least one of the terms must be zero.
   • Either $x - 3 = 0 \implies x = 3$
   • Or $x - 5 = 0 \implies x = 5$

Final Answer: The equation has Two real roots 3, 5.

Question 3: Finding a Repeated Root (from file image_000ddb.png)

The Question: Find the roots of the equation: $x^2 - 2x + 1 = 0$.

Core Concept: The Discriminant and Repeated Roots

Sometimes, a quadratic equation has only one unique solution, which is called a “repeated root”. This happens when the quadratic expression is a perfect square. We can test for this using the discriminant ($\Delta$), which is the part under the square root in the quadratic formula: $\Delta = b^2 - 4ac$.

• If $\Delta > 0$: Two distinct real roots.
• If $\Delta = 0$: One repeated real root.
• If $\Delta < 0$: No real roots.

Detailed Solution:

Method 1: Factoring

1. Analyze the equation: $x^2 - 2x + 1 = 0$.
2. We are looking for two numbers that multiply to +1 and add to -2. The only pair is (-1) and (-1).
3. Rewrite in factored form:
   • $(x - 1)(x - 1) = 0$, which is the same as $(x - 1)^2 = 0$.
4. Solve for the root:
   • The only way for this to be true is if $x - 1 = 0$.
   • $x = 1$.
   • Since the factor $(x-1)$ appears twice, this is a repeated root.

Method 2: Using the Discriminant

1. Identify coefficients from $x^2 - 2x + 1 = 0$:
   • $a = 1$, $b = -2$, $c = 1$.
2. Calculate the discriminant:
   • $\Delta = b^2 - 4ac = (-2)^2 - 4(1)(1) = 4 - 4 = 0$.
3. Interpret the result:
   • Since the discriminant is 0, we know there is exactly one repeated real root. We can find it using the simplified quadratic formula for this case: $x = \frac{-b}{2a} = \frac{-(-2)}{2(1)} = \frac{2}{2} = 1$.

Final Answer: The equation has One repeated, real root 1.