Question

Solve the equations. 3x%2B40%5C%5C%5Cfrac%7Bx%5E%7B2%7D%20-1%7D%7B2%7D%20-11x%3D11" id="TexFormula1" title="3(x+4)^{2} =10x+32\\(x+4)^{2} =3x+40\\\frac{x^{2} -1}{2} -11x=11" alt="3(x+4)^{2} =10x+32\\(x+4)^{2} =3x+40\\\frac{x^{2} -1}{2} -11x=11" align="absmiddle" class="latex-formula">

Answer 1

Answer:

  a) x = {-2 2/3, -2}

  b) x = {-8, 3}

  c) x = {-1, 23}

Step-by-step explanation:

If all you want are solutions, a graphing calculator can give them to you easily. The attachment shows the solutions as x-intercepts when the equation is rearranged to the form f(x) = 0.

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In general, it can be convenient to write the equation in standard form with integer coefficients. Common factors among the coefficients should be removed.

a)

  3(x+4)^2 = 10x +32 . . . . . given

  3(x^2 +8x +16) -10x -32 = 0

  3x^2 +14x +16 = 0

To factor this, we're looking for factors of 3·16 that total 14.

  48 = 1·48 = 2·24 = 3·16 = 4·12 = 6·8

The last pair of factors has a total of 14, so we can rewrite the equation as ...

  (3x +6)(3x +8)/3 = 0

  (x +2)(3x +8) = 0

The solutions are the values of x that make the factors zero:

  x = -2, x = -8/3

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b)

  (x +4)^2 = 3x +40 . . . . . . . . given

  x^2 +8x +16 -3x -40 = 0 . . . . subtract the right side

  x^2 +5x -24 = 0 . . . . . . . . simplify

  (x +8)(x -3) = 0 . . . . . . factor

The solutions are the values of x that make the factors zero:

  x = -8, x = 3

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c)

  (x^2 -1)/2 -11x = 11 . . . . . given

  x^2 -1 -22x -22 = 0 . . . . . . multiply by 2, subtract the right side

  x^2 -22x -23 = 0 . . . . . . simplify

  (x -23)(x +1) = 0 . . . . . . factor

  x = 23, x = -1

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Additional comment

Factoring often gets to the solution with the least fuss when the solutions are rational. There are several ways factoring can be done when the leading coefficient is not 1. One of them is illustrated in (a) above. We will show two other methods that give the same result.

factoring by pairs

We have identified the factors of 3·16 = 48 that have a total of 14. We can use those factors to rewrite the 14x term in the equation.

  3x^2 +6x +8x +16 = 0

Now, we can group the terms in pairs, and factor each pair. It does not matter which of the factors (6 or 8) you write first. You will end with the same result.

  (3x^2 +6x) +(8x +16) = 0

  3x(x +2) +8(x +2) = 0

  (3x +8)(x +2) = 0   ⇒   x = -8/3, x = -2

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factoring using the X method

This method is usually shown using a large graphic X. In the top quadrant is written the product of the leading coefficient and the constant: 3·16 = 48.

In the bottom quadrant is written the coefficient of the x-term: 14.

If one or the other, or both, of these top/bottom values is negative, be sure to keep the sign.

The side quadrants are then filled with values that have a product equal to the top number, and a sum equal to the bottom number. (Pay attention to the signs.) Further, in each of those quadrants, the number written is divided by the leading coefficient, and the fraction reduced to lowest terms.

Here, we would have 6/3 = 2 on one side, and 8/3 on the other side. Now the factors are written as (bx+a), where the reduced fraction on either side is a/b. In our example, the factors are (x+2) and (3x+8)

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alternate X method

This starts off in the same way the X method does, as described above. However, the factors on either side of the X are not divided by the leading coefficient (a). If those side values are 'p' and 'q', the quadratic is written in factored form as ...

  (ax +p)(ax +q)/a = 0

You will notice we used this form above: (3x +6)(3x +8)/3 = 0.

Now, the divisor 'a' can be used to reduce either or both of the numerator factors. Here, the entire factor of 3 can be removed from (3x+6) to make the factorization be ...

  (x +2)(3x +8) = 0

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Sometimes, one factor of the divisor will be removed from one term, and the other factor of it will be removed from the other term. An example of this might be ...

  $\dfrac{(6x+3)(6x+2)}{6}=\dfrac{6x+3}{3}\cdot\dfrac{6x+2}{2}=(2x+1)(3x+1)$