Supply the numbers to write x2+10x+16=0

Question

in its factored form. Enter the lesser number first.(1 point)
(x+

)(x+

)=0
Remaining Attempts : 3

Answer 1

To factor the quadratic equation \( x^2 + 10x + 16 = 0 \), we need to find two numbers that multiply to \( 16 \) (the constant term) and add to \( 10 \) (the coefficient of \( x \)).

The numbers that meet these criteria are \( 2 \) and \( 8 \) because:

• \( 2 \times 8 = 16 \)
• \( 2 + 8 = 10 \)

Thus, we can write the factored form as:

\[ (x + 2)(x + 8) = 0 \]

So the numbers to enter are:

(x + 2)(x + 8) = 0

You should enter:

• The lesser number first:
• \( 2 \)
• \( 8 \)

Final answer:

(x + 2)(x + 8) = 0

Answer 2

Supply the number so that the quadratic equation below has 6 and 8 as its solutions.(1 point)
x2−14x+
=0
Remaining Attempts : 3

Answer 3

To construct a quadratic equation with given solutions (roots) of \(6\) and \(8\), we can use the fact that if \(r_1\) and \(r_2\) are the roots of the quadratic equation \(x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0\), then:

1. The sum of the roots \( (r_1 + r_2) \) is equal to \( 6 + 8 = 14 \).
2. The product of the roots \( (r_1 \cdot r_2) \) is equal to \( 6 \cdot 8 = 48 \).

Now, substituting these values into the standard quadratic form \( x^2 - (sum)x + (product) = 0 \):

\[ x^2 - 14x + 48 = 0 \]

So, the number to fill in the blank is:

48

Answer 4

Solve the following quadratic equation using the perfect square trinomial pattern: x2+10x+25=0
.(1 point)
x=

Answer 5

To solve the quadratic equation \( x^2 + 10x + 25 = 0 \) using the perfect square trinomial pattern, we first recognize that the left side can be factored.

The equation can be rewritten as:

\[ (x + 5)^2 = 0 \]

Next, we take the square root of both sides:

\[ x + 5 = 0 \]

Now, we solve for \( x \):

\[ x = -5 \]

Thus, the solution to the equation is:

\[ x = -5 \]

Answer 6

Solve the following quadratic equation using the perfect square trinomial pattern: x2+54x=−729
.(1 point)
x=

Answer 7

To solve the quadratic equation \( x^2 + 54x = -729 \) using the perfect square trinomial pattern, we first move all terms to one side of the equation:

\[ x^2 + 54x + 729 = 0 \]

Next, we want to write the left-hand side as a perfect square trinomial. We can express \( x^2 + 54x + 729 \) as a squared binomial.

To complete the square, we take half of the coefficient of \( x \) (which is \( 54 \)) and square it:

\[ \left( \frac{54}{2} \right)^2 = 27^2 = 729 \]

So, we can rewrite the equation as:

\[ (x + 27)^2 = 0 \]

Now, take the square root of both sides:

\[ x + 27 = 0 \]

Solving for \( x \):

\[ x = -27 \]

Thus, the solution to the equation is:

\[ x = -27 \]

Answer 8

What are the solutions to x2−121=0
? Enter the lesser number first.(1 point)
x=
, x=

Answer 9

To solve the equation \( x^2 - 121 = 0 \), we can first add \( 121 \) to both sides:

\[ x^2 = 121 \]

Next, we take the square root of both sides:

\[ x = \pm \sqrt{121} \]

Calculating the square root:

\[ \sqrt{121} = 11 \]

Therefore, the solutions are:

\[ x = 11 \quad \text{and} \quad x = -11 \]

Entering the lesser number first, the solutions are:

\[ x = -11, \quad x = 11 \]