5(2X-1)-X²=7X(2-X)

Understand the Problem

The question is asking us to solve the equation 5(2X-1)-X²=7X(2-X) for the variable X. To solve it, we will first expand both sides of the equation, then simplify and combine like terms to form a quadratic equation, and finally use factoring or the quadratic formula to find the values of X.

Answer

$$X_1 = \frac{2 + \sqrt{34}}{6}, \quad X_2 = \frac{2 - \sqrt{34}}{6}$$
Answer for screen readers

The solutions to the equation are:

$$X_1 = \frac{2 + \sqrt{34}}{6}, \quad X_2 = \frac{2 - \sqrt{34}}{6}$$

Steps to Solve

  1. Expand both sides of the equation

Start by distributing the terms on both sides. We have:

$$5(2X - 1) - X^2 = 7X(2 - X)$$

Expanding both sides gives:

$$10X - 5 - X^2 = 14X - 7X^2$$

  1. Rearranging the equation

Next, get all terms on one side of the equation. Move everything to the left side:

$$10X - 5 - X^2 - 14X + 7X^2 = 0$$

This simplifies to:

$$6X^2 - 4X - 5 = 0$$

  1. Simplifying the quadratic equation

Combine like terms:

$$6X^2 - 4X - 5 = 0$$

Now, we can use either factoring or the quadratic formula to solve for $X$.

  1. Applying the quadratic formula

Use the quadratic formula, which is:

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

In our case, $a = 6$, $b = -4$, and $c = -5$. Plugging these values into the formula gives:

$$X = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 6 \cdot (-5)}}{2 \cdot 6}$$

  1. Calculating the values

Calculate the discriminant:

$$(-4)^2 - 4 \cdot 6 \cdot (-5) = 16 + 120 = 136$$

Now substituting back into the formula:

$$X = \frac{4 \pm \sqrt{136}}{12}$$

Since $\sqrt{136} = 2\sqrt{34}$, this simplifies to:

$$X = \frac{4 \pm 2\sqrt{34}}{12} = \frac{2 \pm \sqrt{34}}{6}$$

Which gives us two solutions:

$$X_1 = \frac{2 + \sqrt{34}}{6}, \quad X_2 = \frac{2 - \sqrt{34}}{6}$$

The solutions to the equation are:

$$X_1 = \frac{2 + \sqrt{34}}{6}, \quad X_2 = \frac{2 - \sqrt{34}}{6}$$

More Information

The quadratic formula reveals two possible solutions for this equation. This is common in quadratic equations as they can have zero, one, or two real solutions depending on the discriminant.

Tips

  • Forgetting to combine like terms can lead to an incorrect quadratic equation.
  • Miscalculating the discriminant may give the wrong values for $X$.
  • Not simplifying the final answers can make them look more complicated than necessary.

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