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Solve the following quadratic equation using square roots. Round to the nearest hundredth if necessary: (x+27)^2/−6=−3

why is -23 the wrong answer?
Correct answers:
−22.76, −31.24 And why is this the correct answer

5 answers

To solve the quadratic equation \((x + 27)^2 / -6 = -3\), let's start by isolating the squared term:

  1. Multiply both sides by -6 to eliminate the fraction: \[ (x + 27)^2 = -3 \times -6 \] \[ (x + 27)^2 = 18 \]

  2. Now take the square root of both sides: \[ x + 27 = \pm \sqrt{18} \] Simplifying \(\sqrt{18}\): \[ \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \] (Approximately \(3 \times 1.414 \approx 4.242\))

  3. Now, we have: \[ x + 27 = 3\sqrt{2} \quad \text{or} \quad x + 27 = -3\sqrt{2} \]

  4. Solve for \(x\) by subtracting 27 from both sides in both cases: \[ x = -27 + 3\sqrt{2} \quad \text{or} \quad x = -27 - 3\sqrt{2} \]

  5. Calculate \(x\): For \(x = -27 + 3\sqrt{2}\): \[ x \approx -27 + 4.242 \approx -22.76 \] For \(x = -27 - 3\sqrt{2}\): \[ x \approx -27 - 4.242 \approx -31.24 \]

So the correct answers are approximately \(-22.76\) and \(-31.24\).

Why is \(-23\) the wrong answer?

  • \(-23\) is incorrect because it does not satisfy the original equation. When substituting \(-23\) back in:

    \[ (x + 27)^2 / -6 = -3 \ (-23 + 27)^2 / -6 = -3 \ (4)^2 / -6 \neq -3 \quad \text{(This simplifies to } \frac{16}{-6} = -\frac{8}{3}, \text{ which is not } -3) \]

Thus, \(-23\) is not a valid solution of the equation.

In conclusion, the correct solutions, rounded to two decimal places, are indeed approximately \(-22.76\) and \(-31.24\).

Joana solved the equation 4x^2−4x−3=0
. She followed the steps to solve a quadratic equation by grouping. However, in Step 3 she noticed that the two groupings did not share a common binomial, as shown in her work below. She concluded that the quadratic equation 4x^2−4x−3=0
has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question.

Step 1: 4x^2+2x+(−6x)−3=0


Step 2: (4x^2+2x)+(−6x−3)=0


Step 3: x(4x+2)+(−3(2x+1))=0

where is the shared binomial.

To determine whether Joana's conclusion is correct, let's review her work step-by-step and identify where the issues may have arisen.

Joana's Steps:

Step 1: Joana rewrote the equation as: \[ 4x^2 + 2x - 6x - 3 = 0 \]

Step 2: She grouped the terms as follows: \[ (4x^2 + 2x) + (-6x - 3) = 0 \]

Step 3: She factored out from the groups: \[ x(4x + 2) - 3(2x + 1) = 0 \]

Analysis:

In Step 3, Joana attempted to factor out \(x\) from the first group and \(-3\) from the second group, resulting in the following expression: \[ x(4x + 2) - 3(2x + 1) = 0 \]

The issue arises that \(4x + 2\) and \(2x + 1\) are not the same binomial, which leads to Joana's incorrect conclusion that there is no common binomial shared between the two groups.

Identifying the Common Binomial:

If we analyze the terms she worked with, we can see that:

  1. From the first grouping \(4x^2 + 2x\), the binomial factor is \(2(2x + 1)\).
  2. From the second grouping \(-6x - 3\), the factors with the same terms can also break down to \(-3(2x + 1)\).

Thus during her factoring, both expressions can actually be expressed in terms of \(2x + 1\):

  • \(4x^2 + 2x\) can be factored as \(2(2x + 1)\).
  • \(-6x - 3\) can be factored as \(-3(2x + 1)\).

Step 4 (Correct Factorization):

So a correct grouping should look like this: \[ 2(2x + 1) - 3(2x + 1) = 0 \] Factoring out the common binomial \(2x + 1\): \[ (2x + 1)(2 - 3) = 0 \rightarrow (2x + 1)(-1) = 0 \]

Solving for \(x\):

Setting the factor equal to zero gives: \[ 2x + 1 = 0 \ 2x = -1 \ x = -\frac{1}{2} \]

Conclusion:

Joana's original conclusion that the quadratic equation \(4x^2 - 4x - 3 = 0\) has no solutions is incorrect. The quadratic equation actually has a solution, \(x = -\frac{1}{2}\).

So, the correct conclusion is: the quadratic equation has solutions, and specifically \(x = -\frac{1}{2}\) is one of the solutions.

but where is the shared binomial???

Apologies for any confusion in my previous messages. Let's clarify the identification of the common (or shared) binomial in Joana's work.

Joana’s Original Work

  1. Step 1: \[ 4x^2 + 2x - 6x - 3 = 0 \]

  2. Step 2: \[ (4x^2 + 2x) + (-6x - 3) = 0 \]

  3. Step 3: \[ x(4x + 2) - 3(2x + 1) = 0 \]

Identifying the Shared Binomial

In Step 3, Joana factored the first grouping as \(x(4x + 2)\) and the second grouping as \(-3(2x + 1)\). However, it looks like she made an error in recognizing the common binomial factor.

Observing the expressions:

  • From the first factor, we have \(4x + 2\), which can be factored as \(2(2x + 1)\).
  • From the second factor, we have \(-3(2x + 1)\).

Revealing the Shared Binomial

Now, it is clear that the binomials are:

  • The factor from the first grouping, when correctly factored, is \(2(2x + 1)\).
  • The factor from the second grouping is \(-3(2x + 1)\).

The shared binomial is indeed \((2x + 1)\).

Corrected Step 4 with the Shared Binomial

Now that we know there is a shared binomial, the equation can be rewritten as: \[ (2x + 1)(2 - 3) = 0 \]

Since \(2 - 3 = -1\): \[ (2x + 1)(-1) = 0 \] This simplifies to finding: \[ 2x + 1 = 0 \]

Solve for \(x\):

\[ 2x = -1 \ x = -\frac{1}{2} \]

Conclusion

Thus, to answer your question directly, the shared binomial in Joana's work is \(2x + 1\). This means that the quadratic equation actually does have a solution, specifically \(x = -\frac{1}{2}\). Joana’s error was in not recognizing that there was a common binomial to factor out. Thank you for your patience, and I hope this clears up the confusion!