The Quadratic Formula: Equations with variables on both sides

To solve the quadratic equation $x^2 + 3x - 3 = x$, we first rearrange it to standard form.

Step 1: Move all terms to one side of the equation:
$x^2 + 3x - 3 - x = 0$
This simplifies to:
$x^2 + 2x - 3 = 0$

Step 2: Identify the coefficients in the standard form $ax^2 + bx + c = 0$:
Here, $a = 1$, $b = 2$, and $c = -3$.

Step 3: Use the quadratic formula:
$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$

Plug in the values for $a$, $b$, and $c$:
$x = \frac{{-2 \pm \sqrt{{(2)^2 - 4 \times 1 \times (-3)}}}}{2 \times 1}$

Simplify under the square root:
$x = \frac{{-2 \pm \sqrt{{4 + 12}}}}{2} = \frac{{-2 \pm \sqrt{16}}}{2}$

Simplify further:
$x = \frac{{-2 \pm 4}}{2}$

This results in two solutions:
For the positive square root:
$x = \frac{{-2 + 4}}{2} = \frac{2}{2} = 1$

For the negative square root:
$x = \frac{{-2 - 4}}{2} = \frac{-6}{2} = -3$

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

Since these solutions match choice $x_1 = 1$, $x_2 = -3$, we verify accuracy against the answer choices.

The correct solution to the equation is $x_1 = 1$ and $x_2 = -3$.

To solve the problem $x^4 - x^3 = 2x^2$, let's proceed as follows:

• Step 1: Set the equation to zero.
  $x^4 - x^3 - 2x^2 = 0$
• Step 2: Factor out the greatest common factor.
  The common factor among all terms is $x^2$.
  Factoring out $x^2$ gives:
  $x^2(x^2 - x - 2) = 0$
• Step 3: Solve the factors.
  This equation breaks into two factors that can be solved separately:
  • $x^2 = 0$
  • $x^2 - x - 2 = 0$
• Step 4: Solve $x^2 = 0$.
  Since $x^2 = 0$, we get:
  $x = 0$
• Step 5: Solve $x^2 - x - 2 = 0$.
  This can be factored further. We look for two numbers that multiply to $-2$ and add up to $-1$.
  These numbers are $-2$ and $1$, so we factor as:
  $(x - 2)(x + 1) = 0$
• Step 6: Solve the quadratic factors.
  Set each factor equal to zero:
  • $x - 2 = 0 \Rightarrow x = 2$
  • $x + 1 = 0 \Rightarrow x = -1$

The solutions to the equation $x^4 - x^3 = 2x^2$ are $x = -1, 0, 2$.

Therefore, the correct answer is:

$x = -1, 2, 0$

To solve the equation $3x^2 - 17x + 28 = x + 4$, follow these steps:

• Step 1: Simplify the equation.
  Begin by moving all terms to one side of the equation to obtain a standard form quadratic equation:
  $3x^2 - 17x + 28 - x - 4 = 0$
• Simplify further:
  $3x^2 - 18x + 24 = 0$
• Step 2: Identify coefficients.
  The equation is now in the standard form $ax^2 + bx + c = 0$ where $a = 3$, $b = -18$, and $c = 24$.
• Step 3: Use the quadratic formula to solve for $x$.
  The quadratic formula is given by:
  $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Calculate the discriminant $b^2 - 4ac$:
  $(-18)^2 - 4(3)(24) = 324 - 288 = 36$
• Since the discriminant is positive, there are two solutions.
• Apply the quadratic formula:
  $x = \frac{-(-18) \pm \sqrt{36}}{2(3)} = \frac{18 \pm 6}{6}$
• Calculate the two possible solutions:
  $x_1 = \frac{18 + 6}{6} = 4$
  $x_2 = \frac{18 - 6}{6} = 2$

Therefore, the solutions to the equation are $x_1 = 4$ and $x_2 = 2$.

Let's solve the equation $4x^2 + 9x - 5 = 7 - 4x$ step by step:

• Step 1: Move all terms to one side to form a standard quadratic equation.

We begin by subtracting $7$ and adding $4x$ to both sides of the given equation:

$4x^2 + 9x - 5 - 7 + 4x = 0$

Combining like terms, we get:

$4x^2 + 13x - 12 = 0$

• Step 2: Identify the coefficients $a$, $b$, and $c$ in the quadratic equation $4x^2 + 13x - 12 = 0$.

Here, $a = 4$, $b = 13$, and $c = -12$.

• Step 3: Apply the quadratic formula:

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Plugging in the values of $a$, $b$, and $c$:

$x = \frac{-13 \pm \sqrt{13^2 - 4 \times 4 \times (-12)}}{2 \times 4}$

$x = \frac{-13 \pm \sqrt{169 + 192}}{8}$

$x = \frac{-13 \pm \sqrt{361}}{8}$

$x = \frac{-13 \pm 19}{8}$

• Step 4: Solve for the possible values of $x$.

We have two solutions:

For the positive case:

$x_1 = \frac{-13 + 19}{8} = \frac{6}{8} = \frac{3}{4}$

For the negative case:

$x_2 = \frac{-13 - 19}{8} = \frac{-32}{8} = -4$

Therefore, the solutions are $x_1 = \frac{3}{4}$ and $x_2 = -4$.

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 the problem $x^3 = x^2 + 2x$, follow these steps:

• Step 1: Re-arrange the equation to have all terms on one side:
  $x^3 - x^2 - 2x = 0$.
• Step 2: Factor out the greatest common factor (GCF), which is $x$:
  $x(x^2 - x - 2) = 0$.
• Step 3: Factor the quadratic expression $x^2 - x - 2$:
  The factors of $-2$ that add up to $-1$ are $-2$ and $1$. Thus, $x^2 - x - 2 = (x-2)(x+1)$.
• Step 4: Combine the factored terms:
  $x(x-2)(x+1) = 0$.

Each factor can be set to zero to find the solutions:

• $x = 0$.
• $x - 2 = 0$, so $x = 2$.
• $x + 1 = 0$, so $x = -1$.

The solutions to the equation are $x = 0, -1, 2$.

Therefore, the correct choice from the given options is:
$x = 0, -1, 2$.

To solve the equation $13x^2 + 4x = 8(x+3)^2$, we proceed with the following steps:

• **Step 1: Expand the right-hand side.**
  We start by expanding $8(x+3)^2$:
  $(x+3)^2 = x^2 + 6x + 9$
  Multiplying by 8 gives $8(x^2 + 6x + 9) = 8x^2 + 48x + 72$.
• **Step 2: Form the standard quadratic equation.**
  Now substitute back into the initial equation:
  $13x^2 + 4x = 8x^2 + 48x + 72$
  Rearrange all terms to one side:
  $13x^2 + 4x - 8x^2 - 48x - 72 = 0$
  Simplify: $5x^2 - 44x - 72 = 0$.
• **Step 3: Apply the quadratic formula.**
  The equation is in the standard form $ax^2 + bx + c = 0$ where $a = 5$, $b = -44$, and $c = -72$.
  Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we calculate:
  $x = \frac{-(-44) \pm \sqrt{(-44)^2 - 4 \times 5 \times (-72)}}{2 \times 5}$
  $x = \frac{44 \pm \sqrt{1936 + 1440}}{10}$
  $x = \frac{44 \pm \sqrt{3376}}{10}$
  $\sqrt{3376} \approx 58.1$, so:
  $x = \frac{44 \pm 58.1}{10}$.
• **Step 4: Calculate the roots.**
  For the positive solution:
  $x_1 = \frac{44 + 58.1}{10} = 10.21$.
  For the negative solution:
  $x_2 = \frac{44 - 58.1}{10} = -1.41$.

Therefore, the solutions to the equation are $x_1 = 10.21$ and $x_2 = -1.41$. The correct choice from the given options is choice 3.

To solve the equation $7x^2 + 3x + 8 = 9x + 3$, follow these steps:

• Step 1: Rearrange the equation into standard quadratic form:
  Move all terms to one side:
  $7x^2 + 3x + 8 - 9x - 3 = 0$.
• Step 2: Simplify the equation:
  Combine like terms:
  $7x^2 - 6x + 5 = 0$.
• Step 3: Identify coefficients:
  $a = 7$, $b = -6$, and $c = 5$.
• Step 4: Calculate the discriminant ($\Delta$):
  $\Delta = b^2 - 4ac = (-6)^2 - 4(7)(5) = 36 - 140 = -104$.
• Step 5: Determine the nature of the roots:
  Since the discriminant is negative ($\Delta = -104$), this means there are no real solutions.

Therefore, the solution to the equation is No solution.

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

• Step 1: Simplify the equation by gathering all terms on one side.
• Step 2: Combine like terms to express the equation in standard quadratic form.
• Step 3: Apply the quadratic formula to find the values of $x$.

Now, let's work through each step:

Step 1: Starting with the equation $10x^2 - 65x + 135 = 4x^2 + 13x - 45$, subtract $4x^2$, $13x$, and add $45$ on both sides:

$10x^2 - 65x + 135 - 4x^2 - 13x + 45 = 0$

Step 2: Combine like terms:

$(10x^2 - 4x^2) + (-65x - 13x) + (135 + 45) = 0$

This simplifies to $6x^2 - 78x + 180 = 0$.

Step 3: Identify the coefficients $a = 6$, $b = -78$, and $c = 180$. Use the quadratic formula:

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

Substitute the values:

$x = \frac{-(-78) \pm \sqrt{(-78)^2 - 4 \cdot 6 \cdot 180}}{2 \cdot 6}$

$x = \frac{78 \pm \sqrt{6084 - 4320}}{12}$

$x = \frac{78 \pm \sqrt{1764}}{12}$

$x = \frac{78 \pm 42}{12}$

Calculating the two possible values:

$x_1 = \frac{78 + 42}{12} = \frac{120}{12} = 10$

$x_2 = \frac{78 - 42}{12} = \frac{36}{12} = 3$

Therefore, the solutions to the equation are $x_1 = 10$ and $x_2 = 3$.

The correct answer according to the provided choices is $x_1=10$ and $x_2=3$, which corresponds to choice 4.

To solve this quadratic equation, let us first simplify and rearrange the terms:

Start with the original equation:
$-2x^2 + 6x - 12 = -4x^2 + 19x - 5$

Move all terms to one side to form a standard quadratic equation by adding $4x^2$, subtracting $19x$, and adding $5$ to both sides:

$(-2x^2 + 6x - 12) + 4x^2 - 19x + 5 = 0$

This simplifies to:
$2x^2 - 13x - 7 = 0$

Now, identify the coefficients $a = 2$, $b = -13$, and $c = -7$.

Apply the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Subsitute $a$, $b$, and $c$ into the formula:

$x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \cdot 2 \cdot (-7)}}{2 \cdot 2}$

Simplify:
$x = \frac{13 \pm \sqrt{169 + 56}}{4}$

$x = \frac{13 \pm \sqrt{225}}{4}$

The square root of 225 is 15, thus:

$x = \frac{13 \pm 15}{4}$

Calculate the two possible solutions:

• First solution: $x_1 = \frac{13 + 15}{4} = \frac{28}{4} = 7$
• Second solution: $x_2 = \frac{13 - 15}{4} = \frac{-2}{4} = -\frac{1}{2}$

Therefore, the solutions to the problem are $x_1 = 7$ and $x_2 = -\frac{1}{2}$.

Thus, the correct answer is option 2: $x_1=7$, $x_2=-\frac{1}{2}$.

To solve the equation $-x^2 + x - 2 = -2x^2 - 2x - 4$, we will proceed with the following steps:

• Step 1: Simplify the equation by moving all terms to one side of the equation.
• Step 2: Combine like terms to form a quadratic equation.
• Step 3: Use the quadratic formula to solve for $x$.

Now, let's go through these steps:

Step 1: Start with the given equation:

$-x^2 + x - 2 = -2x^2 - 2x - 4$

Add $2x^2$, $2x$, and $4$ to both sides to move all terms to the left side:

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

Step 2: Combine like terms:

$(2x^2 - x^2) + (x + 2x) + (-2 + 4) = 0$

This simplifies to:

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

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

Calculate the discriminant: $b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 2 = 9 - 8 = 1$.

Since the discriminant is positive, there are two distinct real roots. Substitute into the quadratic formula:

$x = \frac{{-3 \pm \sqrt{1}}}{2 \cdot 1} = \frac{{-3 \pm 1}}{2}$

Calculate the roots:

$x_1 = \frac{{-3 + 1}}{2} = \frac{{-2}}{2} = -1$

$x_2 = \frac{{-3 - 1}}{2} = \frac{{-4}}{2} = -2$

Therefore, the solutions to the equation are $x_1 = -2$ and $x_2 = -1$.

Given the equation:

$x^2 + 3x - 4 = 2x^2$

Step 1: Move all terms to one side:

Subtract $2x^2$ from both sides to get:

$x^2 + 3x - 4 - 2x^2 = 0$

Simplify this to:

$-x^2 + 3x - 4 = 0$

Step 2: Rearrange to standard form:

Multiply the entire equation by -1 for simplicity:

$x^2 - 3x + 4 = 0$

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

Here, $a = 1$, $b = -3$, $c = 4$.

Plug into the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times 4}}{2 \times 1}$

$x = \frac{3 \pm \sqrt{9 - 16}}{2}$

$x = \frac{3 \pm \sqrt{-7}}{2}$

Step 4: Interpret the result:

The discriminant ($b^2 - 4ac$) is negative, $-7$, indicating no real solutions.

Conclusion: The equation has no solution in the set of real numbers.

In comparison with the provided choices, the correct choice is:

No solution

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 $(x-5)^2 - 5 = -12 + 2x$, 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.