The Quadratic Formula: Applying the formula with two solutions

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 solve the quadratic equation $-2X^2 + 6X + 8 = 0$ using the quadratic formula, follow these steps:

• Step 1: Calculate the discriminant, $b^2 - 4ac$.

Here, $a = -2$, $b = 6$, and $c = 8$. Plug these into the formula: $b^2 - 4ac = 6^2 - 4(-2)(8) = 36 + 64 = 100$ Since the discriminant is greater than zero, the roots are real and distinct.

• Step 2: Apply the quadratic formula, $X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Substituting the values, we have: $X = \frac{-6 \pm \sqrt{100}}{2(-2)}$ Simplifying inside the square root gives us: $X = \frac{-6 \pm 10}{-4}$ This leads to two possible solutions: - First, calculate with the positive square root: $X_1 = \frac{-6 + 10}{-4} = \frac{4}{-4} = -1$ - Second, calculate with the negative square root: $X_2 = \frac{-6 - 10}{-4} = \frac{-16}{-4} = 4$

Thus, the solutions to the equation are $X_1 = 4$ and $X_2 = -1$.

Verifying against the choices, the correct choice is: : $X_1=4, X_2=-1$.

Therefore, the solution is $X_1 = 4, X_2 = -1$.

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

To solve the quadratic equation $x^2 + 9x + 8 = 0$, we will use the factoring method because it appears simple to factor.

First, we attempt to factor the quadratic expression $x^2 + 9x + 8$. We look for two numbers that multiply to 8 (the constant term) and add up to 9 (the coefficient of the $x$ term).

These numbers are 1 and 8. So, we can write:

$x^2 + 9x + 8 = (x + 1)(x + 8) = 0$

Now, to find the solutions, we set each factor equal to zero:

1. $x + 1 = 0$ → $x = -1$
2. $x + 8 = 0$ → $x = -8$

Therefore, the solutions to the equation are $x_1 = -1$ and $x_2 = -8$.

Upon reviewing the multiple-choice answers, we find that the correct choice is the one that matches our solutions:

$x_1=-1$ $x_2=-8$

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

To solve the quadratic equation $3x^2 - 39x - 90 = 0$, we will use the quadratic formula.

• Step 1: Identify coefficients: $a = 3$, $b = -39$, and $c = -90$.
• Step 2: Calculate the discriminant: $b^2 - 4ac$.
• Step 3: Apply the quadratic formula to find $x_1$ and $x_2$.

Now, let's work through the steps:

Step 1: Coefficients are given as $a = 3$, $b = -39$, $c = -90$.

Step 2: The discriminant is calculated as follows:

$b^2 - 4ac = (-39)^2 - 4 \cdot 3 \cdot (-90) = 1521 + 1080 = 2601$.

The discriminant is positive, indicating two distinct real solutions.

Step 3: Apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-39) \pm \sqrt{2601}}{2 \cdot 3}$

This simplifies to:

$x = \frac{39 \pm 51}{6}$

Calculating the two solutions:

• $x_1 = \frac{39 + 51}{6} = \frac{90}{6} = 15$.
• $x_2 = \frac{39 - 51}{6} = \frac{-12}{6} = -2$.

Therefore, the solutions to the equation are $x_1 = 15$ and $x_2 = -2$.

Comparing with the choices, the correct answer is:

$x_1 = 15$ $x_2 = -2$

To solve this quadratic equation, we will use the quadratic formula. Let's go through the process step-by-step:

• Step 1: Identify the coefficients.

The coefficients are $a = -2$, $b = 22$, and $c = -60$.

• Step 2: Calculate the discriminant.

The discriminant $D$ is calculated using the formula $b^2 - 4ac$.
Here, $D = 22^2 - 4(-2)(-60) = 484 - 480 = 4$.

• Step 3: Apply the quadratic formula.

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Substituting the values, we get $x = \frac{-22 \pm \sqrt{4}}{2(-2)}$.

• Step 4: Simplify and solve for $x$.

The expression inside the square root is $\sqrt{4} = 2$.
Therefore, we have two potential solutions:
$x_1 = \frac{-22 + 2}{-4} = \frac{-20}{-4} = 5$
$x_2 = \frac{-22 - 2}{-4} = \frac{-24}{-4} = 6$.

The solutions to the equation $-2x^2 + 22x - 60 = 0$ are $x_1 = 5$ and $x_2 = 6$.

In conclusion, the solution to the problem is:

$x_1=5$ $x_2=6$

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 $x = \frac{1}{2}$.

To solve the quadratic equation $4x^2 - 6x - 4 = 0$, we will apply the quadratic formula.

Given the quadratic equation is of the form $ax^2 + bx + c = 0$, we identify:

• $a = 4$
• $b = -6$
• $c = -4$

Next, we use the quadratic formula:

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

Calculate the discriminant:

$b^2 - 4ac = (-6)^2 - 4 \times 4 \times (-4)$

$b^2 - 4ac = 36 + 64 = 100$

Since the discriminant is positive, we have two real solutions.

Now, plug the values into the quadratic formula:

$x = \frac{-(-6) \pm \sqrt{100}}{2 \times 4}$

$x = \frac{6 \pm 10}{8}$

Solving for the two values of $x$:

• First solution: $x_1 = \frac{6 + 10}{8} = \frac{16}{8} = 2$
• Second solution: $x_2 = \frac{6 - 10}{8} = \frac{-4}{8} = -\frac{1}{2}$

Therefore, the solutions to the equation are $x_1 = 2$ and $x_2 = -\frac{1}{2}$.

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

• Step 1: Recognize the given quadratic equation $x^2 - 4x + 4 = 0$ as a perfect square trinomial.
• Step 2: Identify the factored form, which is $(x - 2)^2 = 0$.
• Step 3: Solve the factored equation to find the value of $x$.

Now, let's work through each step:

Step 1: The problem gives us the quadratic equation $x^2 - 4x + 4 = 0$. This equation is a perfect square trinomial because it can be rewritten as $(x - 2)^2 = 0$.

Step 2: Recognize and rewrite the equation in its factored form:

$(x - 2)^2 = 0$.

Step 3: To solve this factored equation, set the factor equal to zero:

$x - 2 = 0$.

Solving for $x$, we get:

$x = 2$.

In this case, the equation has a double root, $x = 2$.

Therefore, the solution to the problem is $x = 2$.

To solve the quadratic equation $5x^2 - 6x + 1 = 0$, we will use the Quadratic Formula:

The Quadratic Formula is given by:

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

Identify the coefficients in the equation:

• $a = 5$
• $b = -6$
• $c = 1$

Step 1: Compute the discriminant $b^2 - 4ac$.

$b^2 - 4ac = (-6)^2 - 4 \times 5 \times 1$

$= 36 - 20$

$= 16$

Step 2: Substitute the values into the Quadratic Formula.

$x = \frac{-(-6) \pm \sqrt{16}}{2 \times 5}$

$= \frac{6 \pm 4}{10}$

Step 3: Calculate the two potential solutions for $x$.

• For $x_1$:

$x_1 = \frac{6 + 4}{10} = \frac{10}{10} = 1$

• For $x_2$:

$x_2 = \frac{6 - 4}{10} = \frac{2}{10} = \frac{1}{5}$

The solutions to the quadratic equation are $x_1 = 1$ and $x_2 = \frac{1}{5}$.

Therefore, after comparing with the provided choices, the correct answer is: $(x_1 = 1, x_2 = \frac{1}{5})$, which matches choice 3.

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.

To solve the quadratic equation $3x^2+10x-8=0$, we use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 3$, $b = 10$, and $c = -8$.

First, calculate the discriminant $b^2 - 4ac$:
$b^2 = 10^2 = 100$
$4ac = 4 \times 3 \times (-8) = -96$
Thus, $b^2 - 4ac = 100 + 96 = 196$.

Since the discriminant (196) is positive, the equation has two distinct real solutions.

Now, substitute into the quadratic formula:
$x = \frac{-10 \pm \sqrt{196}}{2 \times 3} = \frac{-10 \pm 14}{6}$.

This results in two solutions:

• For the plus sign: $x_1 = \frac{-10 + 14}{6} = \frac{4}{6} = \frac{2}{3}$
• For the minus sign: $x_2 = \frac{-10 - 14}{6} = \frac{-24}{6} = -4$

Therefore, the solutions to the given quadratic equation are $x_1 = \frac{2}{3}$ and $x_2 = -4$.

Comparing these results with the multiple-choice options provided, the correct answer is choice 3: $x_1 = \frac{2}{3}, x_2 = -4$.

To solve the equation $-x^2 + 10x - 21 = 0$, we will use the quadratic formula. The equation is in the form $ax^2 + bx + c = 0$ where:

• $a = -1$
• $b = 10$
• $c = -21$

First, compute the discriminant, which is given by $b^2 - 4ac$:

$b^2 - 4ac = (10)^2 - 4(-1)(-21) = 100 - 84 = 16$

Since the discriminant is positive, we have two distinct real solutions. We apply the quadratic formula:

$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-10 \pm \sqrt{16}}}{-2}$

Calculate the roots:

• First root: $x_1 = \frac{{-10 + 4}}{-2} = \frac{{-6}}{-2} = 3$
• Second root: $x_2 = \frac{{-10 - 4}}{-2} = \frac{{-14}}{-2} = 7$

Therefore, the solutions to the equation are $x_1 = 7$ and $x_2 = 3$.

The correct choice from the options provided is:

$x_1=7,x_2=3$

Thus, the solutions to the quadratic equation are $\mathbf{x_1 = 7}$ and $\mathbf{x_2 = 3}$.