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Sophia was solving the following quadratic equation by factoring: `x^{2}-8x-20=0`. Which choice shows the correctly factored form of this equation?

7 answers

To factor the quadratic equation \( x^2 - 8x - 20 = 0 \), we need to find two numbers that multiply to \(-20\) (the constant term) and add up to \(-8\) (the coefficient of the linear term).

The two numbers that satisfy these conditions are \(-10\) and \(2\) because:

  • \( -10 \times 2 = -20 \) (they multiply to \(-20\))
  • \( -10 + 2 = -8 \) (they add up to \(-8\))

Thus, we can factor the quadratic as follows:

\[ x^2 - 8x - 20 = (x - 10)(x + 2) \]

Therefore, the correctly factored form of the equation is:

\[ (x - 10)(x + 2) = 0 \]

Dorothy was solving the following quadratic equation by factoring: `2x^{2}+7x-15=0`. Which choice shows the correctly factored form of this equation?

To factor the quadratic equation \( 2x^2 + 7x - 15 = 0 \), we can use the method of factoring by grouping. We first need to find two numbers that multiply to \( 2 \times (-15) = -30 \) (the product of the leading coefficient and the constant term) and add up to \( 7 \) (the middle coefficient).

The two numbers that multiply to \(-30\) and add up to \(7\) are \(10\) and \(-3\) because:

  • \( 10 \times (-3) = -30 \)
  • \( 10 + (-3) = 7 \)

Next, we can rewrite the middle term of the quadratic equation using these two numbers:

\[ 2x^2 + 10x - 3x - 15 = 0 \]

Now we can group the terms:

\[ (2x^2 + 10x) + (-3x - 15) = 0 \]

Factoring each group gives us:

\[ 2x(x + 5) - 3(x + 5) = 0 \]

Now, we can factor out the common factor \((x + 5)\):

\[ (2x - 3)(x + 5) = 0 \]

Thus, the correctly factored form of the equation is:

\[ (2x - 3)(x + 5) = 0 \]

What is the positive solution to this equation?

`x^{2}+5x\ -\ 36=0`

To find the positive solution to the equation \( x^2 + 5x - 36 = 0 \), we can either factor the equation or use the quadratic formula. Since this equation looks like it can be factored, let's first try factoring.

We want to find two numbers that multiply to \(-36\) (the constant term) and add up to \(5\) (the coefficient of the linear term). The numbers that satisfy this condition are \(9\) and \(-4\) because:

  • \(9 \times (-4) = -36\)
  • \(9 + (-4) = 5\)

Thus, we can factor the quadratic as follows:

\[ x^2 + 5x - 36 = (x + 9)(x - 4) = 0 \]

Now we can set each factor to zero to find the solutions:

\[ x + 9 = 0 \quad \Rightarrow \quad x = -9 \]

\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

The positive solution to the equation \(x^2 + 5x - 36 = 0\) is:

\[ \boxed{4} \]

What is the positive solution to this equation?

`4x^{2}+12x\ =\ 135`

To find the positive solution to the equation \( 4x^2 + 12x = 135 \), we first need to rewrite the equation in standard form by moving all terms to one side:

\[ 4x^2 + 12x - 135 = 0 \]

Next, we can simplify this equation if necessary. We can divide the entire equation by \(4\):

\[ x^2 + 3x - \frac{135}{4} = 0 \]

Now we can use the quadratic formula to solve this equation. The quadratic formula is given by:

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

For our equation \(x^2 + 3x - \frac{135}{4} = 0\), the values are:

  • \(a = 1\)
  • \(b = 3\)
  • \(c = -\frac{135}{4}\)

Substituting these values into the quadratic formula:

  1. Calculate \(b^2 - 4ac\):

\[ b^2 = 3^2 = 9 \]

\[ 4ac = 4 \cdot 1 \cdot \left(-\frac{135}{4}\right) = -135 \]

Thus,

\[ b^2 - 4ac = 9 - (-135) = 9 + 135 = 144 \]

  1. Now substitute back into the formula:

\[ x = \frac{-3 \pm \sqrt{144}}{2 \cdot 1} \]

\[ x = \frac{-3 \pm 12}{2} \]

This gives us two potential solutions:

\[ x = \frac{-3 + 12}{2} = \frac{9}{2} = 4.5 \]

\[ x = \frac{-3 - 12}{2} = \frac{-15}{2} = -7.5 \]

Since we are interested in the positive solution, the positive solution to the equation \(4x^2 + 12x = 135\) is:

\[ \boxed{4.5} \]