Supply the numbers to write x2+10x+16=0 in its factored form. Enter the lesser number first.(1 point) (x+blank)(x+blank)=0

Answer 1

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

The numbers that satisfy these conditions are 2 and 8, since:

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

Thus, we can write the factored form of the equation as:

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

So, the numbers to fill in the blanks are:

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

Entering the lesser number first, you get:

(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+blank=0

Answer 3

To form a quadratic equation with given solutions \(x = 6\) and \(x = 8\), we can use the fact that if \(r_1\) and \(r_2\) are the roots of the equation, the equation can be expressed as:

\[ (x - r_1)(x - r_2) = 0 \]

Substituting in the roots, we have:

\[ (x - 6)(x - 8) = 0 \]

Next, we need to expand this:

\[ x^2 - 8x - 6x + 48 = x^2 - 14x + 48 \]

Thus, the quadratic equation can be written as:

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

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=blank

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 expression \(x^2 + 10x + 25\) can be factored as:

\[ (x + 5)^2 \]

So, we rewrite the equation 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 is:

\(x = -5\).

Answer 6

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

Answer 7

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

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

Next, we focus on the left side. We note that we want to create a perfect square trinomial. The perfect square trinomial has the form \((x + a)^2\).

Completing the Square

1. Identify the coefficient of \(x\), which is 54. Take half of this coefficient and square it: \[ \left(\frac{54}{2}\right)^2 = 27^2 = 729 \]

This confirms we can factor the left side as:

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

2. Now we can write the equation in factored form: \[ (x + 27)^2 = 0 \]

3. Taking the square root on both sides: \[ x + 27 = 0 \]

4. Now, solving for \(x\): \[ x = -27 \]

Thus, the final solution is:

\(x = -27\).

Answer 8

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

Answer 9

To solve the equation \(x^2 - 121 = 0\), we can start by factoring it as a difference of squares:

\[ x^2 - 121 = (x - 11)(x + 11) = 0 \]

Next, we set each factor equal to zero:

1. \(x - 11 = 0\) gives us \(x = 11\)
2. \(x + 11 = 0\) gives us \(x = -11\)

So the solutions are:

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

Thus, entering the lesser number first, the final answer is:

\(x = -11, x = 11\).