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Solve x^2=36

by inspection. There are two real solutions. Enter the lesser number first.(1 point

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11 answers

To solve the equation \( x^2 = 36 \) by inspection, we can find the square roots of 36.

The two real solutions are:

  1. \( x = 6 \)
  2. \( x = -6 \)

Since you asked to enter the lesser number first, the solutions are:

\[ {(-6) (6)} \]

Solve x^2=16/25
by inspection. There are two real solutions. Enter the lesser number first. Leave the answers in simplest fraction form. (

To solve the equation \( x^2 = \frac{16}{25} \) by inspection, we can find the square roots of \(\frac{16}{25}\).

The two real solutions are:

  1. \( x = \sqrt{\frac{16}{25}} = \frac{4}{5} \)
  2. \( x = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \)

Since we need to enter the lesser number first, the solutions are:

\[ \left(-\frac{4}{5} \quad \frac{4}{5}\right) \]

Solve the following quadratic equation using square roots: (x+3)^2=36

To solve the quadratic equation \((x + 3)^2 = 36\) using square roots, follow these steps:

  1. Take the square root of both sides: \[ x + 3 = \pm 6 \]

  2. This gives us two equations to solve:

    • \(x + 3 = 6\)
    • \(x + 3 = -6\)

  3. Solve the first equation: \[ x + 3 = 6 \implies x = 6 - 3 \implies x = 3 \]

  4. Solve the second equation: \[ x + 3 = -6 \implies x = -6 - 3 \implies x = -9 \]

So, the solutions to the equation \((x + 3)^2 = 36\) are:

\[ x = 3 \quad \text{and} \quad x = -9 \]

wrong,To solve this equation, begin by taking the square root of both sides.

Apologies for the confusion earlier. Let's solve the equation \((x + 3)^2 = 36\) step-by-step correctly.

  1. Start by taking the square root of both sides: \[ \sqrt{(x + 3)^2} = \sqrt{36} \] This gives us: \[ x + 3 = \pm 6 \]

  2. Now, we can break this into two separate equations:

    • \(x + 3 = 6\)
    • \(x + 3 = -6\)

  3. Solve the first equation: \[ x + 3 = 6 \implies x = 6 - 3 \implies x = 3 \]

  4. Solve the second equation: \[ x + 3 = -6 \implies x = -6 - 3 \implies x = -9 \]

Therefore, the solutions to the equation \((x + 3)^2 = 36\) are: \[ x = 3 \quad \text{and} \quad x = -9 \]

Thank you for your patience, and I hope this clears up the misunderstanding!

Solve the following quadratic equation using square roots. Round to the nearest hundredth if necessary: (x−6)^2=51
.

To solve the equation \((x - 6)^2 = 51\) using square roots, follow these steps:

  1. Take the square root of both sides: \[ \sqrt{(x - 6)^2} = \sqrt{51} \] This gives: \[ x - 6 = \pm \sqrt{51} \]

  2. Now solve for \(x\) for both cases: \[ x - 6 = \sqrt{51} \quad \text{and} \quad x - 6 = -\sqrt{51} \]

    • For the first equation: \[ x = 6 + \sqrt{51} \]

    • For the second equation: \[ x = 6 - \sqrt{51} \]

  3. Now, calculate \(\sqrt{51}\): \[ \sqrt{51} \approx 7.1414 \quad (\text{to four decimal places}) \]

  4. Substitute this value in the two equations:

    • First solution: \[ x = 6 + 7.1414 \approx 13.1414 \]
    • Second solution: \[ x = 6 - 7.1414 \approx -1.1414 \]

  5. Round to the nearest hundredth:

    • First solution: \[ x \approx 13.14 \]
    • Second solution: \[ x \approx -1.14 \]

Thus, the solutions to the equation \((x - 6)^2 = 51\) are: \[ x \approx 13.14 \quad \text{and} \quad x \approx -1.14 \]

no explanation just the answer

\( x \approx 13.14 \) and \( x \approx -1.14 \)