To solve this rational equation, we first combine the fractions on the left side by finding a common denominator. The common denominator for (1)/(2x-3) and (1)/(x-5) is (2x-3)(x-5). We then add the numerators after adjusting them for the common denominator.

To solve this rational equation, we first combine the fractions on the left side by finding a common denominator. The common denominator for (1)/(2x-3) and (1)/(x-5) is (2x-3)(x-5). We then add the numerators after adjusting them for the common denominator.

Substitute the values of a, b, and c into the quadratic formula and simplify. We calculate the discriminant, √72, which simplifies to 6√2. Finally, we divide all terms by 2 to get the simplified roots.

Substitute the values of a, b, and c into the quadratic formula and simplify. We calculate the discriminant, √72, which simplifies to 6√2. Finally, we divide all terms by 2 to get the simplified roots.

Find the roots of the equation $\frac{1}{2x-3} + \frac{1}{x-5} = 1$ $x \neq \frac{3}{2}, 5$ (rational equation, roots, domain, excluded values)

Answer: The roots of the equation are $x = \frac{8 + 3\sqrt{2}}{2}$ and $x = \frac{8 - 3\sqrt{2}}{2}$ .

Step-by-step solution

Step 1: Combine fractions

To solve this rational equation, we first combine the fractions on the left side by finding a common denominator. The common denominator for $\frac{1}{2x-3}$ and $\frac{1}{x-5}$ is $(2x-3)(x-5)$ . We then add the numerators after adjusting them for the common denominator.

Step 2: Simplify and cross-multiply

Next, we simplify the numerator on the left side. Then, we multiply both sides of the equation by the common denominator $(2x-3)(x-5)$ to eliminate the fractions, which is also known as cross-multiplication.

Step 3: Expand and rearrange into standard quadratic form

We expand the right side of the equation by multiplying the terms. After expanding, we combine like terms and rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$ .

Step 4: Apply the quadratic formula

Since the quadratic equation $2x^2 - 16x + 23 = 0$ cannot be easily factored, we use the quadratic formula to find the roots. Here, $a=2$ , $b=-16$ , and $c=23$ .

Step 5: Substitute values and solve for x

Substitute the values of $a$ , $b$ , and $c$ into the quadratic formula and simplify. We calculate the discriminant, $\sqrt{72}$ , which simplifies to $6\sqrt{2}$ . Finally, we divide all terms by 2 to get the simplified roots.

Step 6: Check for excluded values

The problem states that $x \neq \frac{3}{2}$ (which is 1.5) and $x \neq 5$ . Both calculated roots, approximately $6.12$ and $1.88$ , are not equal to the excluded values. Therefore, both are valid roots.

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Find the roots of the equation 12x−3+1x−5=1\frac{1}{2x-3} + \frac{1}{x-5} = 12x−31​+x−51​=1 x≠32,5x \neq \frac{3}{2}, 5x=23​,5 (rational equation, roots, domain, excluded values)