Using the quadratic formula, find the solution to x2+6x+9=0

Question

.(1 point)

Answer 1

To find the solutions of the quadratic equation $x^2+6x+9=0$ using the quadratic formula, we start by identifying the coefficients $a$, $b$, and $c$:

• $a=1$
• $b=6$
• $c=9$

The quadratic formula is given by:

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

Now, we will calculate the discriminant $b^2-4ac$:

$$b^2=6^2=36$$ $$4ac=4\cdot 1\cdot 9=36$$ $$b^2-4ac=36-36=0$$

Since the discriminant is $0$, we will have one real root. Now substituting into the quadratic formula:

$$x=\frac{-6\pm \sqrt{0}}{2\cdot 1}$$

This simplifies to:

$$x=\frac{-6\pm 0}{2}$$

Thus, we find:

$$x=\frac{-6}{2}=-3$$

Therefore, the solution to the equation $x^2+6x+9=0$ is:

$$\boxed{{\displaystyle -3}}$$

Answer 2

Solve −2x2+3x−1=−2
using the quadratic formula. Round the solution(s) to the nearest hundredths. Enter the smaller solution first.(1 point)

Answer 3

To solve the equation $-2x^2+3x-1=-2$, we first rearrange it into the standard form of a quadratic equation, $a{x}^2+bx+c=0$.

Starting with the equation:

$$-2x^2+3x-1+2=0$$

This simplifies to:

$$-2x^2+3x+1=0$$

Now, we identify the coefficients:

• $a=-2$
• $b=3$
• $c=1$

Next, we apply the quadratic formula:

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

We first calculate the discriminant $b^2-4ac$:

$$b^2=3^2=9$$ $$4ac=4\cdot (-2)\cdot 1=-8$$ $$b^2-4ac=9-(-8)=9+8=17$$

Now substituting back into the quadratic formula:

$$x=\frac{-3\pm \sqrt{17}}{2\cdot -2}$$

This becomes:

$$x=\frac{-3\pm \sqrt{17}}{-4}$$

Now we separate the two possible solutions:

$$x_1=\frac{-3+\sqrt{17}}{-4}$$

$$x_2=\frac{-3-\sqrt{17}}{-4}$$

Calculating the numerical values for these solutions:

1. Calculate $\sqrt{17}$:

   $\sqrt{17}\approx 4.123$

So:

$$x_1=\frac{-3+4.123}{-4}=\frac{1.123}{-4}\approx -0.28075\text{ (rounding gives }-0.28\text{ )}$$

$$x_2=\frac{-3-4.123}{-4}=\frac{-7.123}{-4}\approx 1.78075\text{ (rounding gives }1.78\text{ )}$$

Thus the solutions, rounded to the nearest hundredths, are:

• Smaller solution: $-0.28$
• Larger solution: $1.78$

In conclusion, the solutions are:

$$\boxed{{\displaystyle -0.28}}\text{and}\boxed{{\displaystyle 1.78}}$$

Answer 4

How many real solutions does the following quadratic equation have?

25x2+60x+36=0

Answer 5

To determine how many real solutions the quadratic equation $25x^2+60x+36=0$ has, we need to calculate the discriminant using the formula:

$$D=b^2-4ac$$

where $a$, $b$, and $c$ are the coefficients from the standard form $a{x}^2+bx+c=0$.

For the given equation:

• $a=25$
• $b=60$
• $c=36$

Now, substituting these values into the discriminant formula:

$$D={60}^2-4\cdot 25\cdot 36$$

Calculating ${60}^2$:

$${60}^2=3600$$

Now calculating $4\cdot 25\cdot 36$:

$$4\cdot 25=100$$ $$100\cdot 36=3600$$

Now putting it all together:

$$D=3600-3600=0$$

The value of the discriminant $D$ is $0$.

Conclusion:

When the discriminant is:

• $D>0$: There are two distinct real solutions.
• $D=0$: There is exactly one real solution (a double root).
• $D<0$: There are no real solutions.

Since $D=0$ for this equation, it has exactly one real solution.

Answer 6

Question
Complete the square to solve the following quadratic equation: x2+2x−8=0
.(1 point)
Responses

x=−10
, x=8
x equals negative 10 , x equals 8

x=10
, x=−8
x equals 10 , x equals negative 8

x=2
, x=−4
x equals 2 , x equals negative 4

x=−2
, x=4

Answer 7

To solve the quadratic equation $x^2+2x-8=0$ by completing the square, we will follow these steps:

1. Move the constant term to the other side of the equation:

   $$x^2+2x=8$$

2. Complete the square:

   • Take half of the coefficient of $x$, which is $2$. Half of $2$ is $1$, and squaring it gives $1^2=1$.
   • Add and subtract this square ($1$) inside the equation.

   $$x^2+2x+1-1=8$$

   This simplifies to:

   $$(x+1{)}^2-1=8$$

   Now add $1$ to both sides:

   $$(x+1{)}^2=9$$

3. Take the square root of both sides:

   $$x+1=\pm 3$$

4. Solve for $x$:

   • First solution:

   $$x+1=3⟹x=3-1=2$$

   • Second solution:

   $$x+1=-3⟹x=-3-1=-4$$

Thus, the solutions are:

$$x=2\text{and}x=-4$$

Rounding off to the options given, the correct response is:

$$\boxed{{\displaystyle x=2,;x=-4}}$$

Answer 8

Complete the square to solve the following quadratic equation: x2−22x+1=15 .(1 point)
Responses

x=11+14−−√, 11−14−−√
x equals 11 plus Start Root 14 End Root comma 11 minus Start Root 14 End Root

x=−11+14−−√, −11−14−−√
x equals negative 11 plus Start Root 14 End Root comma negative 11 minus Start Root 14 End Root

x=−11+315−−√, −11−315−−√
x equals negative 11 plus 3 Start Root 15 End Root comma negative 11 minus 3 Start Root 15 End Root

x=11+315−−√, 11−315−−√

Answer 9

To solve the quadratic equation $x^2-22x+1=15$ by completing the square, we will follow these steps:

1. Rearrange the equation: Move $15$ to the left side:

   $$x^2-22x+1-15=0$$

   Simplifying gives:

   $$x^2-22x-14=0$$

2. Move the constant to the other side: $$x^2-22x=14$$

3. Complete the square:

   • Take half of the coefficient of $x$ (which is $-22$), so half of $-22$ is $-11$, and squaring it gives $(-11{)}^2=121$.
   • Add and subtract this square:

   $$x^2-22x+121-121=14$$

   This can be rewritten as:

   $$(x-11{)}^2-121=14$$

   Now add $121$ to both sides:

   $$(x-11{)}^2=14+121=135$$

4. Take the square root of both sides:

   $$x-11=\pm \sqrt{135}$$

5. Solve for $x$:

   $$x=11\pm \sqrt{135}$$

This gives us two solutions:

$$x=11+\sqrt{135}\text{and}x=11-\sqrt{135}$$

Final Answer:

Now, reducing $\sqrt{135}$:

$$\sqrt{135}=\sqrt{9\cdot 15}=3\sqrt{15}$$

So the solutions can also be written as:

$$x=11+3\sqrt{15}\text{and}x=11-3\sqrt{15}$$

The correct response is:

$$\boxed{{\displaystyle x=11+3\sqrt{15},;11-3\sqrt{15}}}$$

Answer 10

Using the quadratic formula, find the solution to x2−6x−16=0 .(1 point)
Responses

x=8, x=−2
x equals 8 , x equals negative 2

x=8
x equals 8

x=−2
x equals negative 2

x=8, x=−8

Answer 11

To solve the quadratic equation $x^2-6x-16=0$ using the quadratic formula, we start by identifying the coefficients $a$, $b$, and $c$:

• $a=1$
• $b=-6$
• $c=-16$

The quadratic formula is given by:

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

Step 1: Calculate the Discriminant

We first calculate the discriminant $D$:

$$D=b^2-4ac$$

Substituting the values of $a$, $b$, and $c$:

$$D=(-6{)}^2-4\cdot 1\cdot (-16)$$ $$D=36+64$$ $$D=100$$

Step 2: Apply the Quadratic Formula

Now substitute the values of $b$, $D$, and $a$ into the quadratic formula:

$$x=\frac{-(-6)\pm \sqrt{100}}{2\cdot 1}$$ $$x=\frac{6\pm 10}{2}$$

Step 3: Solve for $x$

This gives us two potential solutions:

1. For the positive case: $$x_1=\frac{6+10}{2}=\frac{16}{2}=8$$

2. For the negative case: $$x_2=\frac{6-10}{2}=\frac{-4}{2}=-2$$

Final Solutions

Thus, the two solutions to the equation $x^2-6x-16=0$ are:

$$x=8\text{and}x=-2$$

Correct Response

The correct response is:

$$\boxed{{\displaystyle x=8,;x=-2}}$$