The Quadratic Formula: Using short multiplication formulas

To solve the equation $(3x + 1)^2 + 8 = 12$, we start by isolating the squared expression:

• First, subtract 8 from both sides to simplify: $(3x + 1)^2 = 12 - 8$.
• This gives $(3x + 1)^2 = 4$.

Next, we take the square root of both sides to remove the square:

• $3x + 1 = \pm 2$, recognizing the positive and negative roots of 4.

We now solve for $x$ in each case:

• Case 1: $3x + 1 = 2$
  Subtract 1 from both sides: $3x = 1$
  Divide by 3: $x = \frac{1}{3}$.
• Case 2: $3x + 1 = -2$
  Subtract 1 from both sides: $3x = -3$
  Divide by 3: $x = -1$.

Therefore, the solutions to the original equation are $x_1 = \frac{1}{3}$ and $x_2 = -1$.

To solve this quadratic equation, follow the steps below:

• Step 1: Begin by expanding $(x+4)^2$.
• Step 2: Expand to get $(x+4)^2 = x^2 + 8x + 16$.
• Step 3: Substitute the expanded form into the original equation: $7 = 5x^2 + 8x + x^2 + 8x + 16$.\
• Step 4: Combine like terms: $7 = 6x^2 + 16x + 16$.
• Step 5: Rearrange into standard quadratic form: $6x^2 + 16x + 9 = 0$.
• Step 6: Use the quadratic formula $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$, where $a = 6$, $b = 16$, and $c = 9$.
• Step 7: Compute the discriminant: $b^2 - 4ac = 16^2 - 4(6)(9) = 256 - 216 = 40$.
• Step 8: Substitute into the quadratic formula: $x = \frac{{-16 \pm \sqrt{40}}}{12} = \frac{{-16 \pm 2\sqrt{10}}}{12} = -\frac{4}{3} \pm \frac{\sqrt{10}}{6}$.

Thus, the solutions are $x = -\frac{4}{3} + \frac{\sqrt{10}}{6}$ and $x = -\frac{4}{3} - \frac{\sqrt{10}}{6}$.

Therefore, the correct solution, corresponding to the provided choices, is $-\frac{4}{3} \pm \frac{\sqrt{10}}{6}$.

To solve the equation $7x + 1 + (2x + 3)^2 = (4x + 2)^2$, we follow these steps:

• Step 1: Expand both sides using the square of a binomial formula.
• Step 2: Simplify the equation to form a standard quadratic equation.
• Step 3: Use the quadratic formula to find the roots of the equation.

Step 1: Expand the squares.

The left side: $(2x + 3)^2 = 4x^2 + 12x + 9$.

The right side: $(4x + 2)^2 = 16x^2 + 16x + 4$.

Step 2: Substitute back into the original equation and simplify:

$7x + 1 + 4x^2 + 12x + 9 = 16x^2 + 16x + 4$.

Combine like terms:

$4x^2 + 19x + 10 = 16x^2 + 16x + 4$.

Step 3: Move all terms to one side:

$4x^2 + 19x + 10 - 16x^2 - 16x - 4 = 0$.

Which simplifies to:

$-12x^2 + 3x + 6 = 0$.

Step 4: Divide by -3 to simplify:

$4x^2 - x - 2 = 0$.

Step 5: Use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 4$, $b = -1$, $c = -2$.

Calculate the discriminant:

$b^2 - 4ac = (-1)^2 - 4 \cdot 4 \cdot (-2) = 1 + 32 = 33$.

Calculate the roots:

$x = \frac{1 \pm \sqrt{33}}{8}$.

Therefore, the solution to the problem is $x = \frac{1 \pm \sqrt{33}}{8}$.

To solve this equation, we follow these steps:

• Step 1: Multiply both sides by $(x-1)^2$ to eliminate the fraction.
• Step 2: Expand and simplify both sides of the equation.
• Step 3: Rearrange the equation to form a polynomial equal to zero.
• Step 4: Solve the resulting polynomial using factorization or the quadratic formula.

Now, let's execute these steps:

Step 1: Multiply both sides by $(x-1)^2$:
$(x^3 + 1) = (x + 4)(x - 1)^2$

Step 2: Expand the right side:
$(x + 4)(x^2 - 2x + 1) = x(x^2 - 2x + 1) + 4(x^2 - 2x + 1)$

Calculating each part yields:
$x(x^2 - 2x + 1) = x^3 - 2x^2 + x$
$4(x^2 - 2x + 1) = 4x^2 - 8x + 4$

Add these together:
$x^3 - 2x^2 + x + 4x^2 - 8x + 4 = x^3 + 2x^2 - 7x + 4$

Step 3: Combine terms and rearrange:
$x^3 + 1 = x^3 + 2x^2 - 7x + 4$

Simplify by cancelling $x^3$ from both sides:
$1 = 2x^2 - 7x + 4$

Move 1 to the right side:
$0 = 2x^2 - 7x + 3$

Step 4: Solve the quadratic equation $2x^2 - 7x + 3 = 0$.

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -7$, and $c = 3$.

Calculate the discriminant:
$b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot 3 = 49 - 24 = 25$

Now plug into the quadratic formula:
$x = \frac{7 \pm \sqrt{25}}{4}$

Simplify:
$x = \frac{7 \pm 5}{4}$

Two solutions arise:
$x = \frac{12}{4} = 3$ and $x = \frac{2}{4} = \frac{1}{2}$

Since $x = 1$ would make the denominator zero, it is not a valid solution for the original equation.

Therefore, the solution to the problem is $x = 3$ or $x = \frac{1}{2}$.

To solve $-(x+3)^2 = 4x$, follow these steps:

• Step 1: Expand the left side: $-(x+3)^2 = -(x^2 + 6x + 9)$.
• Step 2: Distribute the negative sign: $-x^2 - 6x - 9$.
• Step 3: Set the equation by moving terms to the right: $-x^2 - 6x - 9 = 4x$ becomes $-x^2 - 6x - 9 - 4x = 0$.
• Step 4: Simplify to standard quadratic form: $-x^2 - 10x - 9 = 0$.
• Step 5: Applying the quadratic formula where $a = -1$, $b = -10$, $c = -9$:
• $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot (-1) \cdot (-9)}}{2 \cdot (-1)}$.
• $x = \frac{10 \pm \sqrt{100 - 36}}{-2}$.
• $x = \frac{10 \pm \sqrt{64}}{-2}$.
• $x = \frac{10 \pm 8}{-2}$.
• Two solutions arise: $x = \frac{10 + 8}{-2} = -9$ and $x = \frac{10 - 8}{-2} = -1$.

Therefore, the solutions are $x_1 = -1$ and $x_2 = -9$.

Thus, the correct answer is $\mathbf{x_1=-1,x_2=-9}$.

We will solve the equation $(x+2)^2 = (2x+3)^2$ by expanding and simplifying both sides:

Step 1: Expand both sides of the equation:
Left side: $(x+2)^2 = x^2 + 4x + 4$
Right side: $(2x+3)^2 = 4x^2 + 12x + 9$

Step 2: Set the expanded forms equal to each other:
$x^2 + 4x + 4 = 4x^2 + 12x + 9$

Step 3: Rearrange to form a standard quadratic equation:
Subtract $x^2 + 4x + 4$ from both sides:
$0 = 3x^2 + 8x + 5$

Step 4: Rearrange to get:
$3x^2 + 8x + 5 = 0$

Step 5: Solve using the quadratic formula:
Using $a = 3$, $b = 8$, $c = 5$:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3}$
$x = \frac{-8 \pm \sqrt{64 - 60}}{6}$
$x = \frac{-8 \pm \sqrt{4}}{6}$
$x = \frac{-8 \pm 2}{6}$

Step 6: Calculate the solutions:
$x_1 = \frac{-8 + 2}{6} = \frac{-6}{6} = -1$
$x_2 = \frac{-8 - 2}{6} = \frac{-10}{6} = -\frac{5}{3}$

Verify in the original equation to assure correctness. Hence, both solutions are valid.

Therefore, the solutions are $x_1 = -1$ and $x_2 = -\frac{5}{3}$, which matches choice 3.

To solve the given equation, follow these steps:

• Step 1: Expand $(x - 4)^2$ using the formula $(a - b)^2 = a^2 - 2ab + b^2$.

Thus, $(x - 4)^2 = x^2 - 8x + 16$.

• Step 2: Substitute the expanded form into the equation:

$x^2 - 8x + 16 + 3x^2 = -16x + 12$.

• Step 3: Combine like terms on the left-hand side.

This gives $4x^2 - 8x + 16 = -16x + 12$.

• Step 4: Rearrange the equation to set it to zero.

Bring all terms to one side: $4x^2 - 8x + 16 + 16x - 12 = 0$.

Combine and simplify the terms: $4x^2 + 8x + 4 = 0$.

• Step 5: Simplify the equation by dividing each term by 4.

It becomes $x^2 + 2x + 1 = 0$.

• Step 6: Recognize the equation as a perfect square trinomial.

$(x + 1)^2 = 0$.

• Step 7: Solve by taking the square root of both sides.

The solution is $x + 1 = 0$, therefore $x = -1$.

In conclusion, the solution to the equation is $x = -1$.

To solve the equation $(x+3)^2 + 2x^2 = 18$, we'll follow these steps:

• Step 1: Expand the expression $(x+3)^2$.
• Step 2: Combine and simplify terms to form a standard quadratic equation.
• Step 3: Use the quadratic formula to find values of $x$.

Now, let's work through each step.
Step 1: Expand $(x+3)^2$:

$(x+3)^2 = x^2 + 2 \times x \times 3 + 3^2 = x^2 + 6x + 9$.

Step 2: Substitute back into the original equation:

$x^2 + 6x + 9 + 2x^2 = 18$.

Combine like terms:

$3x^2 + 6x + 9 = 18$.

Subtract 18 from both sides to form the quadratic equation:

$3x^2 + 6x + 9 - 18 = 0$.

Simplify:

$3x^2 + 6x - 9 = 0$.

Divide every term by 3 to simplify further:

$x^2 + 2x - 3 = 0$.

Step 3: Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=2$, and $c=-3$.

Calculate discriminant: $b^2 - 4ac = 2^2 - 4 \times 1 \times (-3) = 4 + 12 = 16$.

Since the discriminant is positive, there are two real roots.

Find roots:

$x = \frac{-2 \pm \sqrt{16}}{2 \times 1}$.

Calculate roots:

$x_1 = \frac{-2 + 4}{2} = 1$,
$x_2 = \frac{-2 - 4}{2} = -3$.

Therefore, the solutions to the equation are $x_1 = 1$, $x_2 = -3$.

To solve the equation $(x-5)^2 - 5 = -12 + 2x$, follow these steps:

• Step 1: Expand the square on the left side of the equation:
  $(x-5)^2 = x^2 - 10x + 25$
• Step 2: Substitute this back into the equation:
  $x^2 - 10x + 25 - 5 = -12 + 2x$
• Step 3: Simplify the equation:
  $x^2 - 10x + 20 = -12 + 2x$
• Step 4: Rearrange the equation by moving all terms to one side:
  $x^2 - 10x + 20 - 2x + 12 = 0$
  This simplifies to $x^2 - 12x + 32 = 0$.
• Step 5: Use the Quadratic Formula, where $a = 1$, $b = -12$, and $c = 32$:
  $x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \times 1 \times 32}}{2 \times 1}$
• Step 6: Calculate the discriminant and simplify:
  $x = \frac{12 \pm \sqrt{144 - 128}}{2}$
  $x = \frac{12 \pm \sqrt{16}}{2}$
  $x = \frac{12 \pm 4}{2}$
• Step 7: Solve for the two potential values of $x$:
  $x_1 = \frac{12 + 4}{2} = 8$
  $x_2 = \frac{12 - 4}{2} = 4$

Thus, the solutions to the equation are $x_1 = 8$ and $x_2 = 4$.

Therefore, the correct answer is $x_1 = 8, x_2 = 4$, which corresponds to choice 1.

To solve the equation $(x+3)^2 = 2x + 5$, we proceed as follows:

• Step 1: Expand the left side. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, we find:
  $(x+3)^2 = x^2 + 6x + 9$.

• Step 2: Set the equation to zero by moving all terms to one side:
  $x^2 + 6x + 9 = 2x + 5$
  Subtract $2x + 5$ from both sides:
  $x^2 + 6x + 9 - 2x - 5 = 0$
  This simplifies to:
  $x^2 + 4x + 4 = 0$.

• Step 3: Solve the quadratic equation $x^2 + 4x + 4 = 0$. Notice this can be factored as:
  $(x+2)^2 = 0$.

• Step 4: Solve for $x$ by setting the factor equal to zero:
  $x+2 = 0$.
  Thus, $x = -2$.

Therefore, the solution to the equation $(x+3)^2 = 2x + 5$ is $x = -2$.

The given equation is:

$2x^2 - 2x = (x+1)^2$

Step 1: Expand the right-hand side.

$(x+1)^2 = x^2 + 2x + 1$

Step 2: Write the full equation with the expanded form.

$2x^2 - 2x = x^2 + 2x + 1$

Step 3: Bring all terms to one side of the equation to set it to zero.

$2x^2 - 2x - x^2 - 2x - 1 = 0$

Step 4: Simplify the equation.

$x^2 - 4x - 1 = 0$

Step 5: Identify coefficients for the quadratic formula.

Here, $a = 1$, $b = -4$, $c = -1$.

Step 6: Apply the quadratic formula.

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}$

$x = \frac{4 \pm \sqrt{16 + 4}}{2}$

$x = \frac{4 \pm \sqrt{20}}{2}$

$x = \frac{4 \pm 2\sqrt{5}}{2}$

$x = 2 \pm \sqrt{5}$

Therefore, the solutions are $x = 2 + \sqrt{5}$ and $x = 2 - \sqrt{5}$.

These solutions correspond to choice (4): Answers a + b

To solve this problem, we'll follow these steps:

• Step 1: Cross-multiply the given equation to eliminate fractions.
• Step 2: Expand the squared terms on either side of the equation.
• Step 3: Rearrange terms to form a quadratic equation.
• Step 4: Solve the quadratic equation to find possible values for $x$.

Now, let's work through each step:

Step 1: Begin with the given equation:
$\frac{(\frac{1}{x} + \frac{1}{2})^2}{(\frac{1}{x} + \frac{1}{3})^2} = \frac{81}{64}$. Cross-multiply to eliminate fractions:
$(\frac{1}{x} + \frac{1}{2})^2 \times 64 = (\frac{1}{x} + \frac{1}{3})^2 \times 81$.

Step 2: Expand each squared term:
For $(\frac{1}{x} + \frac{1}{2})^2$, use $(a + b)^2 = a^2 + 2ab + b^2$:
$(\frac{1}{x})^2 + 2(\frac{1}{x})(\frac{1}{2}) + (\frac{1}{2})^2 = \frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}$.
Similarly, $(\frac{1}{x} + \frac{1}{3})^2 = \frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}$.

Step 3: Substitute these into the cross-multiplied equation:
$64\left(\frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}\right) = 81\left(\frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}\right)$.

Step 4: Simplify and collect like terms:
$64(\frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}) = (64\frac{1}{x^2} + 64\frac{1}{x} + 16)$,
$81(\frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}) = (81\frac{1}{x^2} + 54\frac{1}{x} + 9)$.

Equating terms gives:
$64\frac{1}{x^2} + 64\frac{1}{x} + 16 = 81\frac{1}{x^2} + 54\frac{1}{x} + 9$.

Step 5: Solve the quadratic equation:
Combine like terms: $-17\frac{1}{x^2} + 10\frac{1}{x} + 7 = 0$.
Let $y = \frac{1}{x}$. Substitute to get: $-17y^2 + 10y + 7 = 0$.
Multiply the entire equation by -1 to simplify: $17y^2 - 10y - 7 = 0$.

Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=17$, $b=-10$, $c=-7$:
$y = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 17 \times (-7)}}{2 \times 17}$
$y = \frac{10 \pm \sqrt{100 + 476}}{34}$
$y = \frac{10 \pm \sqrt{576}}{34}$
$y = \frac{10 \pm 24}{34}$
Which gives:
$y = \frac{34}{34} = 1$ or $y = -\frac{14}{34} = -\frac{7}{17}$.

Since $y = \frac{1}{x}$:
For $y=1$, $x = \frac{1}{y} = 1$.
For $y=-\frac{7}{17}$, $x = \frac{1}{y} = -\frac{17}{7}$.

Therefore, the solutions for $x$ are $x = 1$ and $x = -\frac{17}{7}$.

Checking the correct answer choice, these correspond to the second choice.

Thus, the solution to the problem is $x = 1, -\frac{17}{7}$.

To solve the equation $\frac{x^3 + 1}{(x+1)^2} = x$, we will follow these steps:

• Step 1: Set up the equation for solving by cross-multiplying.
• Step 2: Simplify and solve the resulting polynomial equation.
• Step 3: Solve for the values of $x$.

Let's work through the solution:

Step 1: Cross-multiply to eliminate the fraction:

Expand the right-hand side:

Step 2: Set the expanded equation equal:

Cancel $x^3$ from both sides:

Re-arrange the equation to form a standard quadratic equation:

Step 3: Solve the quadratic equation using the quadratic formula:

The quadratic formula is:

Substitute the values of $a$, $b$, and $c$ into the formula:

Calculate the discriminant and simplify:

Simplify further:

This gives the solutions:

Since $x = -1$ would make the denominator zero, it is not allowed as a solution. Thus, the only valid solution is:

Therefore, the solution to the equation is $x = \frac{1}{2}$.

To solve the given equation $(x-5)^2 - 5 = 10 + 2x$, we'll follow these steps:

• Step 1: Expand and simplify the left side of the equation.
• Step 2: Move all terms to form a standard quadratic equation.
• Step 3: Use the quadratic formula to find the values of $x$.

Now, let's work through each step:

Step 1: Expand the left side.
$(x-5)^2 = x^2 - 10x + 25$
The equation becomes:
$x^2 - 10x + 25 - 5 = 10 + 2x$

Step 2: Collect all terms on one side.
$x^2 - 10x + 20 = 10 + 2x$
Subtract $10 + 2x$ from both sides to get:
$x^2 - 10x + 20 - 10 - 2x = 0$
This simplifies to:
$x^2 - 12x + 10 = 0$

Step 3: Apply the quadratic formula:
For $ax^2 + bx + c = 0$, the formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a = 1$, $b = -12$, $c = 10$.
Calculate the discriminant:
$b^2 - 4ac = (-12)^2 - 4 \cdot 1 \cdot 10 = 144 - 40 = 104$
Now, solve for $x$:
$x = \frac{-(-12) \pm \sqrt{104}}{2 \cdot 1} = \frac{12 \pm \sqrt{104}}{2}$

Therefore, the solutions to the equation are:
$x_1 = 6 + \frac{\sqrt{104}}{2}$, $x_2 = 6 - \frac{\sqrt{104}}{2}$.

This matches the correct choice, confirming that the solution is correct.

In order to solve the equation, start by removing the denominators.

To do this, we'll multiply the denominators:

$(2x+1)^2\cdot(2x+1)+(x+2)^2\cdot(x+2)=4.5x(2x+1)(x+2)$

Open the parentheses on the left side, making use of the distributive property:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x+1)(x+2)$

Continue to open the parentheses on the right side of the equation:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x^2+5x+2)$

Simplify further:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=9x^3+22.5x+9x$

Go back and simplify the parentheses on the left side of the equation:

$8x^3+8x^2+2x+4x^2+4x+1+x^3+4x^2+4x+2x^2+8x+8=9x^3+22.5x+9x$

Combine like terms:

$9x^3+18x^2+18x+9=9x^3+22.5x+9x$

Notice that all terms can be divided by 9 as shown below:

$x^3+2x^2+2x+1=x^3+2.5x+x$

Move all numbers to one side:

$x^3-x^3+2x^2-2.5x^2+2x-x+9=0$

We obtain the following:

$0.5x^2-x-1=0$

In order to remove the one-half coefficient, multiply the entire equation by 2

$x^2-2x-2=0$

Apply the square root formula, as shown below-

$\frac{2±\sqrt{12}}{2}$

Apply the properties of square roots in order to simplify the square root of 12:

$\frac{2±2\sqrt{3}}{2}$Divide both the numerator and denominator by 2 as follows:

$1±\sqrt{3}$

To solve this problem, we'll follow these steps:

• Step 1: Clear fractions by multiplying through by the least common denominator.
• Step 2: Simplify the resulting equation.
• Step 3: Solve the quadratic equation using the quadratic formula.

Now, let's work through each step:
**Step 1:** Multiply both sides by $(x+1)^2$ to clear the denominators:
$(x+1)^2 \left(\frac{1}{(x+1)^2} + \frac{1}{x+1}\right) = (x+1)^2 \cdot 1$
This simplifies to:
$1 + (x+1) = (x+1)^2$
**Step 2:** Simplify the equation:
$1 + x + 1 = x^2 + 2x + 1$
Combine like terms:
$2 + x = x^2 + 2x + 1$
Rearrange to form a quadratic equation:
$x^2 + 2x + 1 - x - 2 = 0$
Thus, we have:
$x^2 + x - 1 = 0$
**Step 3:** Solve the quadratic equation $x^2 + x - 1 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 1$, and $c = -1$.
Calculate the discriminant:
$b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-1) = 1 + 4 = 5$
Thus, $x$ is:
$x = \frac{-1 \pm \sqrt{5}}{2}$
**Conclusion:** The solutions to the equation are:
$x = \frac{-1 + \sqrt{5}}{2}$ and $x = \frac{-1 - \sqrt{5}}{2}$
Upon verifying with given choices, the correct answer is:
$x = -\frac{1}{2}[1\pm\sqrt{5}\rbrack$