Solving Quadratic Equations - Examples, Exercises and Solutions

Let's recall the quadratic formula:

We'll substitute the given data into the formula:

$x={{-(-10)\pm\sqrt{-10^2-4\cdot2\cdot(-12)}\over 2\cdot2}}$

Let's simplify and solve the part under the square root:

$x={{10\pm\sqrt{100+96}\over 4}}$

$x={{10\pm\sqrt{196}\over 4}}$

$x={{10\pm14\over 4}}$

Now we'll solve using both methods, once with the addition sign and once with the subtraction sign:

$x={{10+14\over 4}} = {24\over4}=6$

$x={{10-14\over 4}} = {-4\over4}=-1$

We've arrived at the solution: X=6,-1

Let's solve the problem step-by-step:

First, consider the given equation:

$x^2 + 7x = 0$

This equation is almost in the standard form of a quadratic equation:

$ax^2 + bx + c = 0$

Where:

• $a$ is the coefficient of $x^2$
• $b$ is the coefficient of $x$
• $c$ is the constant term (independent number)

Let's identify each of these components from the given equation:

• For $a$: The term with $x^2$ is $x^2$. In this case, the coefficient is implicitly 1, so $a = 1$.
• For $b$: The term with $x$ is $7x$. The coefficient of $x$ is 7, so $b = 7$.
• For $c$: There is no independent constant term visible, so we assume $c = 0$.

Thus, the components of the quadratic equation are:

$a = 1$, $b = 7$, $c = 0$

The correct choice from the provided options is : $a=1$, $b=7$, $c=0$

Let's solve this problem step-by-step by identifying the coefficients of the quadratic equation:

First, examine the given equation:

$5 - 6x^2 + 12x = 0$

To make it easier to identify the coefficients, we rewrite the equation in the standard quadratic form:

$-6x^2 + 12x + 5 = 0$

In this expression, we can now directly identify the coefficients:

• The coefficient of $x^2$ (quadratic term) is $a = -6$.
• The coefficient of $x$ (linear term) is $b = 12$.
• The constant term (independent number) is $c = 5$.

Thus, the components of the quadratic equation are:

$a = -6$, $b = 12$, $c = 5$

By comparing these values to the multiple-choice options, we can determine that the correct choice is:

Choice 4: $a = -6$, $b = 12$, $c = 5$

Therefore, the final solution is:

$a = -6$, $b = 12$, $c = 5$.

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.

This is a quadratic equation:

$x^2+3x-18=0$

This is 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 a form where all terms on one side are ordered from highest to 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 the 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:

$x^2+3x-18=0$

And solve it:

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

$\begin{cases}a=1\\b=3\\c=-18\end{cases}$

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

And we'll obtain the solutions of the equation (its roots) by substituting the coefficients we just noted in the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-3\pm\sqrt{3^2-4\cdot1\cdot(-18)}}{2\cdot1}$

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

$x_{1,2}=\frac{-3\pm\sqrt{81}}{2}=\frac{-3\pm9}{2}$

Therefore the solutions of the equation are:

$$$\begin{cases}x_1=\frac{-3+9}{2}=3 \\ x_2=\frac{-3-9}{2}=-6\end{cases}$

Therefore the correct answer is answer C.

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.

The given quadratic equation is $x^2 + 10x + 25 = 0$.

Notice that $x^2 + 10x + 25$ is a perfect square trinomial, which can be factored as $(x + 5)^2$. Let's verify by expanding:

$(x + 5)^2 = (x + 5)(x + 5) = x^2 + 5x + 5x + 25 = x^2 + 10x + 25$.

Since the factoring is correct, we rewrite the equation:

$(x + 5)^2 = 0$.

Take the square root of both sides:

$x + 5 = 0$.

Solving for $x$, we find:

$x = -5$.

The equation has a double root, meaning the solution is repeated. Thus, the final solution is:

$x = -5$.

This matches the given correct answer.

The parameters are expressed in the quadratic equation as follows:

aX2+bX+c=0

We substitute into the formula:

-5±√(5²-4*1*4) 
          2

-5±√(25-16)
         2

-5±√9
    2

-5±3
   2

The symbol ± means that we have to solve this part twice, once with a plus and a second time with a minus,

This is how we later get two results.

-5-3 = -8
-8/2 = -4

-5+3 = -2
-2/2 = -1

And thus we find out that X = -1, -4

The quadratic equation we have is $x^2 + 4x - 5 = 0$.

We'll compare this with the general form of a quadratic equation: $ax^2 + bx + c = 0$.

1. Identify $a$: The coefficient of $x^2$ in the given equation is $1$. Therefore, $a = 1$.

2. Identify $b$: The coefficient of $x$ in the given equation is $4$. Therefore, $b = 4$.

3. Identify $c$: The constant term in the given equation is $-5$. Therefore, $c = -5$.

Thus, the components of the equation are:

• $a = 1$
• $b = 4$
• $c = -5$

The correct answer to this problem, matching choice id 3, is:

$a=1$ $b=4$ $c=-5$

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 answer the question, we'll need to recall the quadratic formula:

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

Let's remember that:

a is the coefficient of X²

b is the coefficient of X

c is the free term

And if we look again at the formula given to us:

a=1

b=10

c=9

Let's substitute into the formula:

$x = {-10 \pm \sqrt{10^2-4\cdot 1 \cdot 9} \over 2\cdot 1}$

Let's start by solving what's under the square root:

$x = {-10 \pm \sqrt{100-36} \over 2}$

$x = {-10 \pm \sqrt{64} \over 2}$

$x = {-10 \pm 8 \over 2}$

Now we'll solve twice, once with plus and once with minus

$x = {-10 +8 \over 2}= {-2 \over 2} = -1$

$x = {-10 -8 \over 2} = {-18 \over 2} =-9$

And we can see that we got two solutions, X=-1 and X=-9

And that's the solution!

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 $a = -1$.

To determine the components of the quadratic equation, follow these steps:

• Step 1: Recognize the standard form of a quadratic equation, which is $ax^2 + bx + c = 0$.
• Step 2: Compare the given equation $10x^2 + 20x + 5 = 0$ to the standard form.
• Step 3: Identify the coefficients:
  - The term $10x^2$ indicates that $a = 10$.
  - The term $20x$ indicates that $b = 20$.
  - The term $5$ is the constant term, so $c = 5$.

Therefore, the components of the equation are:

$a = 10$, $b = 20$, $c = 5$.

The correct answer among the choices provided is the one that correctly identifies these coefficients:

$a=10$ $b=20$ $c=5$

Therefore, the correct choice is Choice 4.

To identify the coefficients from the quadratic equation $5x^2 + 6x - 8 = 0$, follow these steps:

• Standard Form of Quadratic Equation: Compare $5x^2 + 6x - 8$ with the standard form $ax^2 + bx + c = 0$.
• Identify $a$, $b$, and $c$:
  • $a = 5$: The coefficient of $x^2$.
  • $b = 6$: The coefficient of $x$.
  • $c = -8$: The constant term (independent of $x$).

Therefore, from the equation $5x^2 + 6x - 8 = 0$, the coefficients are identified as $a=5$, $b=6$, and $c=-8$.

Comparing with choices, we find that choice 2 is correct: $a=5$, $b=6$, $c=-8$.

Thus, the coefficients are identified as $a=5$, $b=6$, $c=-8$.