- #1

Albert1

- 1,221

- 0

$(\dfrac {13x-x^2}{x+1})(x+\dfrac{13-x}{x+1})=42$

$find \,\, real \,\, x$

$find \,\, real \,\, x$

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- MHB
- Thread starter Albert1
- Start date

In summary, to find the real x value in a mathematical equation, you can use algebraic methods such as factoring, completing the square, or using the quadratic formula. A calculator can also be used to find the real x value, and in the case of multiple x values, plugging in each value into the original equation can help determine the real x value. However, equations with fractions, square roots, or absolute value signs may require additional techniques. The "guess and check" method may not be the most reliable option for finding the real x value.

- #1

Albert1

- 1,221

- 0

$(\dfrac {13x-x^2}{x+1})(x+\dfrac{13-x}{x+1})=42$

$find \,\, real \,\, x$

$find \,\, real \,\, x$

Mathematics news on Phys.org

- #2

lfdahl

Gold Member

MHB

- 749

- 0

\[\left ( \frac{13x-x^2}{x+1} \right )\left ( x+\frac{13-x}{x+1} \right )=\left ( \frac{13x-x^2}{x+1} \right ) \left ( \frac{13+x^2}{x+1} \right )=42 \\\\ \Rightarrow \left (14-\left (x+\frac{14}{x+1} \right ) \right )\left ( -1+\left ( x+\frac{14}{x+1} \right ) \right )=42\]Let $x+\frac{14}{x+1} = \alpha$ and solve the quadratic equation:\[(14-\alpha )(-1+\alpha )=42\Rightarrow -\alpha^2+15\alpha -56 = 0\Rightarrow \alpha \in \left \{ \frac{15\pm 1}{2} \right \}=\left \{ 7,8 \right \}\]Using the $\alpha$-expression, we get:

\[x+\frac{14}{x+1}=7\Rightarrow x^2-6x+7 = 0\Rightarrow x\in \left \{ 3\pm \sqrt{2} \right \} \\\\ x+\frac{14}{x+1}=8 \Rightarrow x^2-7x+6 = 0 \Rightarrow x\in \left \{ 1,6 \right \}\]

- thus the set of solutions is: \[S = \left \{ 1,3-\sqrt{2},3+\sqrt{2},6 \right \}\]

To find the real x value, you will need to solve the equation by using algebraic methods such as factoring, completing the square, or using the quadratic formula.

Yes, you can use a calculator to find the real x value. Most scientific or graphing calculators have functions that can solve equations for x.

If the equation has multiple x values, you will need to plug in each value into the original equation and see which one satisfies the equation. The value that makes the equation true is the real x value.

Yes, there are special cases such as equations with fractions, square roots, or absolute value signs. In these cases, you may need to use additional algebraic techniques or follow specific rules to find the real x value.

While the "guess and check" method may work in some cases, it is not the most efficient or reliable method for finding the real x value. It is best to use algebraic methods or a calculator to accurately find the real x value.

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