Solve Equations of the Quadratic Form

Solve the equation:

$$ 0.1x^4 - 1.3x^2 + 3.6 = 0 $$

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Solution:

Let $u = x^2$ which gives $u^2 = x^4$, and rewrite the given equation in terms of $u$:

$$ 0.1u^2 - 1.3u + 3.6 = 0 $$

Solve the above quadratic equation to find $u$.

$$ u = 4 \quad \text{and} \quad u = 9 $$

We now use the substitution $u = x^2$ to solve for $x$.

$$ x^2 = 4 \implies x = \pm 2 $$

$$ x^2 = 9 \implies x = \pm 3 $$

The four x-intercepts of the graph of $y = 0.1x^4 - 1.3x^2 + 3.6$ are the graphical solutions to the equation as shown below.

Solve the equation:

$$ \sqrt{x} = 3 - \dfrac{1}{4}x $$

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Solution:

Let $u = \sqrt{x}$ which gives $u^2 = x$, and rewrite the equation in terms of $u$:

$$ u = 3 - \dfrac{u^2}{4} $$

Multiply all terms by 4, simplify, write the quadratic in standard form and solve:

$$ 4u = 12 - u^2 $$

$$ u^2 + 4u - 12 = 0 $$

$$ (u + 6)(u - 2) = 0 $$

$$ u = -6 \quad \text{and} \quad u = 2 $$

Use the substitution $u = \sqrt{x}$ to solve for $x$.

$$ \sqrt{x} = -6 \implies \text{no real solution} $$

$$ \sqrt{x} = 2 \implies x = 4 $$

The x-intercept of the graph is the graphical solution to the equation as shown below.

Solve the equation:

$$ \left(3 - \dfrac{4}{x}\right)^2 - 6\left(3 - \dfrac{4}{x}\right) = 16 $$

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Solution:

Let $y = 3 - \dfrac{4}{x}$ which gives $y^2 = \left(3 - \dfrac{4}{x}\right)^2$, and rewrite the equation in terms of $y$:

$$ y^2 - 6y = 16 $$

Simplify and solve:

$$ y^2 - 6y - 16 = 0 $$

$$ (y - 8)(y + 2) = 0 $$

$$ y = -2 \quad \text{and} \quad y = 8 $$

Solve for $x$.

For $y = -2$, we substitute back into the equation for $x$:

$$ 3 - \dfrac{4}{x} = -2 \implies \dfrac{4}{x} = 5 \implies x = \dfrac{4}{5} $$

For $y = 8$, we substitute back into the equation for $x$:

$$ 3 - \dfrac{4}{x} = 8 \implies \dfrac{4}{x} = -5 \implies x = -\dfrac{4}{5} $$

The x-intercepts of the graph are the graphical solutions to the equation as shown below.

Solve the equation:

$$ 2(x - 1)^{2/3} + 3(x - 1)^{1/3} - 2 = 0 $$

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Solution:

Let $y = (x - 1)^{1/3}$ which gives $y^2 = (x - 1)^{2/3}$, and rewrite the equation in terms of $y$:

$$ 2y^2 + 3y - 2 = 0 $$

$$ (2y - 1)(y + 2) = 0 $$

$$ y = -2 \quad \text{and} \quad y = \dfrac{1}{2} $$

Solve for $x$.

$$ (x - 1)^{1/3} = -2 \implies x - 1 = (-2)^3 \implies x - 1 = -8 \implies x = -7 $$

$$ (x - 1)^{1/3} = \dfrac{1}{2} \implies x - 1 = \left(\dfrac{1}{2}\right)^3 \implies x - 1 = \dfrac{1}{8} \implies x = \dfrac{9}{8} $$

The x-intercepts of the graph are the graphical solutions to the equation as shown below.

Find all real solutions for the equation:

$$ 2\left(\dfrac{2}{x - 3}\right)^2 - \dfrac{2}{x - 3} - 3 = 0 $$

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Solution:

Let $u = \dfrac{2}{x - 3}$ which gives $u^2 = \left(\dfrac{2}{x - 3}\right)^2$, and rewrite the equation in terms of $u$:

$$ 2u^2 - u - 3 = 0 $$

$$ (2u - 3)(u + 1) = 0 $$

$$ u = -1 \quad \text{and} \quad u = \dfrac{3}{2} $$

Solve for $x$.

$$ \dfrac{2}{x - 3} = -1 \implies x - 3 = -2 \implies x = 1 $$

$$ \dfrac{2}{x - 3} = \dfrac{3}{2} \implies 3(x - 3) = 4 \implies 3x - 9 = 4 \implies 3x = 13 \implies x = \dfrac{13}{3} $$

The x-intercepts of the graph are the graphical solutions to the equation as shown below.

Solve the equation for all real values of $x$:

$$ x^{-2} - 3x^{-1} - 4 = 0 $$

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Solution:

This equation is in quadratic form if we let $u = x^{-1}$ (which means $u = \dfrac{1}{x}$). Notice that $u^2 = (x^{-1})^2 = x^{-2}$.

Rewrite the equation in terms of $u$:

$$ u^2 - 3u - 4 = 0 $$

Factor the quadratic:

$$ (u - 4)(u + 1) = 0 $$

This gives us two solutions for $u$:

$$ u = 4 \quad \text{and} \quad u = -1 $$

Now, substitute $x^{-1}$ back in for $u$ to solve for $x$:

For $u = 4$:

$$ \dfrac{1}{x} = 4 \implies x = \dfrac{1}{4} $$

For $u = -1$:

$$ \dfrac{1}{x} = -1 \implies x = -1 $$

Both solutions are valid since $x \neq 0$.

Final Answers: $x = \dfrac{1}{4}, \ x = -1$

Find the exact real solutions for the exponential equation:

$$ e^{2x} - 5e^x + 6 = 0 $$

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Solution:

Using the properties of exponents, recognize that $e^{2x} = (e^x)^2$. This means the equation is quadratic in form.

Let $u = e^x$. The equation becomes:

$$ u^2 - 5u + 6 = 0 $$

Factor the quadratic equation:

$$ (u - 2)(u - 3) = 0 $$

This yields the roots:

$$ u = 2 \quad \text{and} \quad u = 3 $$

Now substitute back $u = e^x$:

For $u = 2$:

$$ e^x = 2 \implies \ln(e^x) = \ln(2) \implies x = \ln(2) $$

For $u = 3$:

$$ e^x = 3 \implies \ln(e^x) = \ln(3) \implies x = \ln(3) $$

Since the exponential function $e^x$ is always positive, both $u=2$ and $u=3$ are valid (no extraneous roots are produced).

Final Answers: $x = \ln(2), \ x = \ln(3)$