Solving Quadratic Equations - Examples, Exercises and Solutions

Let's recall the general form of the quadratic function:

$y=ax^2+bx+c$ The function given in the problem is:

$y=-x^2+25x$$c$is the free term (meaning the coefficient of the term with power 0),

In the function in the problem there is no free term,

Therefore, we can identify that:

$c=0$Therefore, the correct answer is answer A.

To solve this problem, we'll follow these steps:

• Identify the given information and the standard quadratic form
• Compare the given equation to this standard form
• Extract the coefficients $a$, $b$, and $c$ and find $b$

Now, let's work through each step:
Step 1: The problem provides us with the equation $y = 4x^2 - 16$. It's already in a form where we can identify the coefficients.
Step 2: Recall the standard form of a quadratic equation is $ax^2 + bx + c$. Compare this form to the equation $y = 4x^2 - 16$.
Step 3: By comparison, the coefficient of $x^2$ (which is $a$) is 4. There is no $x$ term explicitly present, implying that $b = 0$. The constant $c$ is -16.
Therefore, after comparison and identification, it becomes clear that the value of $b$ in the equation is $b = 0$.

To solve this problem, we'll follow these steps:

• Step 1: Compare the given equation $y = 5 + 3x^2$ to the standard form $ax^2 + bx + c$.
• Step 2: Identify the terms corresponding to $a$, $b$, and $c$.

Now, let's work through each step:
Step 1: The given equation is $y = 5 + 3x^2$. Rearranging it in the standard form, we have $y = 3x^2 + 0\cdot x + 5$.

Step 2: From this arrangement, it's clear that:
- $a = 3$ (the coefficient of $x^2$)
- $b = 0$ (there is no $x$ term, so its coefficient is 0)
- $c = 5$ (the constant term)

Therefore, the value of $c$ is $\ c=5$.

To solve this problem, we need to identify the coefficient of $x$ in the given quadratic equation. The equation given is $y = 3x^2 + 10 - x$. Let’s rearrange this equation to match the standard form of a quadratic equation $ax^2 + bx + c$.

The given equation can be rewritten as:

$y = 3x^2 - x + 10$

Here, we can identify the coefficients:

• $a = 3$ (for $x^2$)
• $b = -1$ (for $x$)
• $c = 10$ (the constant term)

Therefore, the value of $b$, the coefficient of $x$, is $-1$.

To solve this problem, we'll follow these steps:

• Step 1: Rewrite the given equation in standard quadratic form if necessary.
• Step 2: Identify the term with $x^2$ in the equation.
• Step 3: Extract the coefficient of $x^2$ as $a$.

Now, let's work through each step:
Step 1: The provided equation is $y = 3x - 10 + 5x^2$. Although it's not initially in standard form, observation shows that the $x^2$ term is clearly present.
Step 2: Locate the $x^2$ term: in our equation, this term is $5x^2$.
Step 3: The coefficient of $x^2$ is $5$. Hence, $a = 5$.

Therefore, the coefficient of $x^2$, or $a$, is $a = 5$.

To solve this problem, we'll follow these steps:

• Identify the given quadratic equation.
• Compare the given equation with the standard quadratic form $ax^2 + bx + c$.
• Determine the value of $c$ by direct comparison.

Now, let's work through each step:
Step 1: The given quadratic equation is $y = -5x^2 + 4x - 3$.
Step 2: The standard form of a quadratic equation is $ax^2 + bx + c$. We need to match the coefficients accordingly.
Step 3: By comparing the terms from the equation with the standard form, $a$ is the coefficient of $x^2$, $b$ is the coefficient of $x$, and $c$ is the constant term or the independent number.

Therefore, from the equation $y = -5x^2 + 4x - 3$:

• $a = -5$
• $b = 4$
• $c = -3$

Thus, the value of $c$ in the quadratic equation is $c = -3$.

To solve this problem, we need to identify the coefficients in the given quadratic equation:

• The equation provided is $y = 2x - 3x^2 + 1$.
• The standard form of a quadratic equation is $ax^2 + bx + c$.
• From the equation, identify:
  • The $ax^2$ term is $-3x^2$, indicating $a = -3$.
  • The $bx$ term is $2x$, indicating $b = 2$.
  • The constant term $c$ is $1$.

Thus, the coefficient $b$ in the equation $y = 2x - 3x^2 + 1$ is $b = 2$, which corresponds to choice 1.

To determine the coefficient $a$ in the given quadratic equation $y = -x^2 - 3x + 1$, follow these steps:

• Step 1: Recognize the form of the quadratic equation as $y = ax^2 + bx + c$.
• Step 2: Identify the $x^2$ term in the equation $y = -x^2 - 3x + 1$.
• Step 3: Determine the coefficient of the $x^2$ term, which is in front of $x^2$.

In the equation $y = -x^2 - 3x + 1$, the term involving $x^2$ is $-x^2$, where the coefficient $a$ is clearly $-1$.

Hence, the value of $a$ is $a = -1$.

This is a quadratic equation:

$x^2+5x+4=0 =0$

due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in to a form where all terms on one side are ordered from the highest to the lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed to solve it using the quadratic formula.

Remember:

The rule states that the roots of an equation of the form:

$ax^2+bx+c=0$

are:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

$x^2+5x+4=0 =0$

And solve it:

First, let's identify the coefficients of the terms:

$\begin{cases}a=1 \\ b=5 \\ c=4\end{cases}$

where we noted that the coefficient of the quadratic term is 1,

We obtain the solutions of the equation (its roots) by insertion we just identified into the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot4}}{2\cdot1}$

Let's continue to calculate the expression inside of the square root and simplify the expression:

$x_{1,2}=\frac{-5\pm\sqrt{9}}{2}=\frac{-5\pm3}{2}$

Therefore the solutions of the equation are:

$$$\begin{cases}x_1=\frac{-5+3}{2}=-1 \\ x_2=\frac{-5-3}{2}=-4\end{cases}$

Therefore the correct answer is answer C.

This is a quadratic equation:

$x^2+5x+6=0$

due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it to a form where all the terms on one side are ordered from the highest to the lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed to solve it using the quadratic formula,

Remember:

The rule states that the roots of an equation of the form:

$ax^2+bx+c=0$

are:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

$x^2+5x+6=0$and solve it:

First, let's identify the coefficients of the terms:

$\begin{cases}a=1 \\ b=5 \\ c=6\end{cases}$

where we noted that the coefficient of the quadratic term is 1,

We obtain the equation's solutions (roots) by inserting the coefficients we just noted into the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot6}}{2\cdot1}$

Let's continue to calculate the expression inside of the square root and proceed to simplify the expression:

$x_{1,2}=\frac{-5\pm\sqrt{1}}{2}=\frac{-5\pm1}{2}$

The solutions to the equation are:

$$$\begin{cases}x_1=\frac{-5+1}{2}=-2 \\ x_2=\frac{-5-1}{2}=-3\end{cases}$

Therefore the correct answer is answer D.

To solve the quadratic equation $x^2 - 3x + 2 = 0$, we'll follow these steps:

• Step 1: Check for factorization. Assuming the quadratic can be factored, we look for two numbers that multiply to $c = 2$ and add to $b = -3$. These numbers are $-1$ and $-2$.
• Step 2: Factor the quadratic as $(x - 1)(x - 2) = 0$.
• Step 3: Solve each factor for $x$.

Now, let's solve the factors:
From $(x - 1) = 0$, we have $x = 1$.
From $(x - 2) = 0$, we have $x = 2$.

Thus, the solutions to the equation are $x_1 = 1$ and $x_2 = 2$.

Therefore, the solution to the problem is $x_1 = 1, x_2 = 2$.

To solve the quadratic equation $x^2 - x - 20 = 0$ using the quadratic formula, follow these steps:

• Step 1: Identify the coefficients $a$, $b$, and $c$ in the equation. For our equation $x^2 - x - 20 = 0$, we have $a = 1$, $b = -1$, and $c = -20$.
• Step 2: Substitute these values into the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Step 3: Calculate the discriminant $\Delta = b^2 - 4ac$:
  $\Delta = (-1)^2 - 4 \cdot 1 \cdot (-20) = 1 + 80 = 81$
• Step 4: Since the discriminant is positive, there are two distinct real roots. Substitute back into the quadratic formula:
  $x = \frac{-(-1) \pm \sqrt{81}}{2 \cdot 1} = \frac{1 \pm 9}{2}$
• Step 5: Solve for the two possible values of $x$:
  $x_1 = \frac{1 + 9}{2} = 5$ $x_2 = \frac{1 - 9}{2} = -4$

Therefore, the solutions to the equation $x^2 - x - 20 = 0$ are $x_1 = 5$ and $x_2 = -4$.

Accordingly, the correct choice matches with $x_1 = -4, x_2 = 5$, which is option 3.

The given equation is:

$x^2 - 4x + 4 = 0$

This resembles a perfect square trinomial. The expression $x^2 - 4x + 4$ can be rewritten as $(x-2)^2$. This can be verified by expanding $(x-2)(x-2)$ to confirm it equals $x^2 - 4x + 4$.

Therefore, the equation becomes:

$(x-2)^2 = 0$

To solve for $x$, take the square root of both sides:

$x - 2 = 0$

Adding 2 to both sides gives:

$x = 2$

Thus, the solution to the equation $x^2 - 4x + 4 = 0$ is $x = 2$, which corresponds to the unique real root of the equation.

To solve this quadratic equation $x^2 - 2x - 3 = 0$, we will employ the quadratic formula.

• Step 1: Identify the coefficients: $a = 1$, $b = -2$, and $c = -3$.
• Step 2: Calculate the discriminant $\Delta = b^2 - 4ac$.
• Step 3: Substitute into the quadratic formula to find the roots.

Now, let's work through each step:

Step 1: The coefficients are $a = 1$, $b = -2$, $c = -3$.

Step 2: Calculate the discriminant:
$\Delta = (-2)^2 - 4 \times 1 \times (-3) = 4 + 12 = 16$.

Step 3: Substitute into the quadratic formula:
$x = \frac{-(-2) \pm \sqrt{16}}{2 \times 1} = \frac{2 \pm 4}{2}$.

This gives us two solutions:

• For the '+' sign: $x_1 = \frac{2 + 4}{2} = \frac{6}{2} = 3$.
• For the '-' sign: $x_2 = \frac{2 - 4}{2} = \frac{-2}{2} = -1$.

Therefore, the solutions to the equation $x^2 - 2x - 3 = 0$ are $x_1 = 3$ and $x_2 = -1$, which corresponds to choice 2.

To solve the equation $4x^2 - 4x + 1 = 0$, we will use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, we identify $a = 4$, $b = -4$, and $c = 1$.

Calculate the discriminant:

$b^2 - 4ac = (-4)^2 - 4 \times 4 \times 1 = 16 - 16 = 0$

Since the discriminant is 0, there is one real repeated root.

Substitute into the quadratic formula:

$x = \frac{-(-4) \pm \sqrt{0}}{2 \times 4} = \frac{4}{8} = \frac{1}{2}$

Therefore, the solution to the equation is $x = \frac{1}{2}$.