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Solve the quadratic equation 46x2−138x=0

by factoring out the GCF.(1 point)
Responses

The solutions are x=0
and x=3
.
The solutions are x equals 0 and x equals 3 .

The solutions are x=0
and x=46
.
The solutions are x equals 0 and x equals 46 .

The solutions are x=−46
and x=3
.
The solutions are x equals negative 46 and x equals 3 .

The solutions are x=−3
and x=46
.

11 answers

To solve the quadratic equation \(46x^2 - 138x = 0\) by factoring out the greatest common factor (GCF), follow these steps:

  1. Identify the GCF of the terms, which is \(46x\).
  2. Factor out \(46x\):

\[ 46x(x - 3) = 0 \]

  1. Set each factor equal to zero:

\[ 46x = 0 \quad \text{or} \quad x - 3 = 0 \]

  1. Solve the equations:

\[ x = 0 \quad \text{or} \quad x = 3 \]

Thus, the solutions to the equation are:

The solutions are \(x = 0\) and \(x = 3\).

Use factoring to find the solutions for 30=−x2+13x
.(1 point)
Responses

x=−10,−3
x equals negative 10 comma negative 3

x=3,10
x equals 3 comma 10

x=−3,10
x equals negative 3 comma 10

x=−10,3

To solve the equation \(30 = -x^2 + 13x\) by factoring, first rearrange the equation to standard quadratic form:

\[ 0 = -x^2 + 13x - 30 \]

Now, we can rewrite this as:

\[ x^2 - 13x + 30 = 0 \]

Next, we need to factor the quadratic. We are looking for two numbers that multiply to \(30\) (the constant term) and add to \(-13\) (the coefficient of \(x\)). The two numbers that fit this are \(-3\) and \(-10\).

So, we can factor the equation as:

\[ (x - 3)(x - 10) = 0 \]

Now, set each factor equal to zero:

  1. \(x - 3 = 0\) → \(x = 3\)
  2. \(x - 10 = 0\) → \(x = 10\)

Thus, the solutions to the equation are:

The solutions are \(x = 3\) and \(x = 10\).

The correct response is:

x = 3, 10 or \(x\) equals \(3\) comma \(10\).

Solve the following quadratic equation using the perfect square trinomial pattern: 49x2=−392x−784
.(1 point)
Responses

x=−4
x equals negative 4

x=4
x equals 4

x=−784
x equals negative 784

x=784

To solve the quadratic equation \(49x^2 = -392x - 784\) using the perfect square trinomial pattern, we first rearrange the equation to bring all terms to one side:

\[ 49x^2 + 392x + 784 = 0 \]

Next, we need to check if it can be expressed as a perfect square trinomial. A perfect square trinomial can be written in the form \((ax + b)^2\).

To do this, we start by checking if it can be factored:

  1. The first term \(49x^2\) can be written as \((7x)^2\).
  2. The last term \(784\) can be written as \(28^2\).
  3. Now we need to see if the middle term \(392x\) fits the pattern. The middle term in the expansion of \((7x + 28)^2\) is \(2 \cdot 7x \cdot 28\).

Calculating this:

\[ 2 \cdot 7 \cdot 28 = 392 \]

So, we can factor the left-hand side:

\[ (7x + 28)^2 = 0 \]

Now, we set the factor equal to zero:

\[ 7x + 28 = 0 \]

Solving for \(x\):

\[ 7x = -28 \ x = -4 \]

Thus, the solution to the equation is:

x = -4 or \(x\) equals negative 4.

Grant was asked to solve 49x2−225=0
by factoring using the difference of squares pattern. What is/are the correct solution(s)?(1 point)

The given equation is:

\[ 49x^2 - 225 = 0 \]

This equation is a difference of squares, which can be factored using the formula \(a^2 - b^2 = (a - b)(a + b)\).

  1. Identify \(a\) and \(b\):

    • \(a^2 = 49x^2\) implies \(a = 7x\)
    • \(b^2 = 225\) implies \(b = 15\) (since \(15^2 = 225\))

  2. Use the difference of squares factorization:

\[ 49x^2 - 225 = (7x - 15)(7x + 15) = 0 \]

  1. Set each factor equal to zero:

\[ 7x - 15 = 0 \quad \text{or} \quad 7x + 15 = 0 \]

  1. Solve for \(x\):

From \(7x - 15 = 0\):

\[ 7x = 15 \ x = \frac{15}{7} \]

From \(7x + 15 = 0\):

\[ 7x = -15 \ x = -\frac{15}{7} \]

Thus, the solutions to the equation \(49x^2 - 225 = 0\) are:

\[ x = \frac{15}{7} \quad \text{and} \quad x = -\frac{15}{7} \]

These are the correct solutions.

Complete the square to solve the following quadratic equation: x2+4x=41
.(1 point)
Responses

x=2+53–√, 2−53–√
x equals 2 plus 5 Start Root 3 End Root comma 2 minus 5 Start Root 3 End Root

x=−2+53–√, −2−53–√
x equals negative 2 plus 5 Start Root 3 End Root comma negative 2 minus 5 Start Root 3 End Root

x=2+35–√, 2−35–√
x equals 2 plus 3 Start Root 5 End Root comma 2 minus 3 Start Root 5 End Root

x=−2+35–√, −2−35–√

To solve the quadratic equation \(x^2 + 4x = 41\) by completing the square, follow these steps:

  1. Rearrange the equation: \[ x^2 + 4x - 41 = 0 \] (or directly work with \(x^2 + 4x = 41\)).

  2. Move the constant term to the right side: \[ x^2 + 4x = 41 \]

  3. To complete the square, take half of the coefficient of \(x\), square it, and add it to both sides. The coefficient of \(x\) is \(4\), so half of it is \(2\), and squaring gives \(2^2 = 4\).

  4. Add \(4\) to both sides of the equation: \[ x^2 + 4x + 4 = 41 + 4 \] Simplifying gives: \[ (x + 2)^2 = 45 \]

  5. Take the square root of both sides: \[ x + 2 = \pm \sqrt{45} \]

  6. Simplify \(\sqrt{45}\): \[ \sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5} \]

  7. Now we have: \[ x + 2 = \pm 3\sqrt{5} \]

  8. Finally, solve for \(x\): \[ x = -2 \pm 3\sqrt{5} \]

Thus, the solutions are:

\[ x = -2 + 3\sqrt{5} \quad \text{and} \quad x = -2 - 3\sqrt{5} \]

The correct response is:

\(x = -2 + 3\sqrt{5}, -2 - 3\sqrt{5}\).

Using the quadratic formula, find the solution to 4x2+4x−15=0
.(1 point)
Responses

x=1.5
x equals 1.5

x=−1.5
, x=−2.5
x equals negative 1.5 , x equals negative 2.5

x=1.5
, x=−2.5
x equals 1.5 , x equals negative 2.5

x=−2.5

To solve the quadratic equation \(4x^2 + 4x - 15 = 0\) using the quadratic formula, we will use the formula:

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

where \(a\), \(b\), and \(c\) are the coefficients from the standard form \(ax^2 + bx + c = 0\).

For the equation \(4x^2 + 4x - 15 = 0\):

  • \(a = 4\)
  • \(b = 4\)
  • \(c = -15\)

Now, we will substitute these values into the quadratic formula:

  1. Calculate the discriminant \(b^2 - 4ac\): \[ b^2 - 4ac = 4^2 - 4(4)(-15) = 16 + 240 = 256 \]

  2. Plug the values into the quadratic formula: \[ x = \frac{-4 \pm \sqrt{256}}{2 \cdot 4} \] \[ x = \frac{-4 \pm 16}{8} \]

  3. Calculate the two possible values for \(x\):

    • For the plus case: \[ x = \frac{-4 + 16}{8} = \frac{12}{8} = \frac{3}{2} = 1.5 \]

    • For the minus case: \[ x = \frac{-4 - 16}{8} = \frac{-20}{8} = -\frac{5}{2} = -2.5 \]

Thus, the solutions are:

\[ x = 1.5 \quad \text{and} \quad x = -2.5 \]

The correct response is:

\(x = 1.5, x = -2.5\) or \(x\) equals \(1.5\), \(x\) equals \(-2.5\).

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