Question:
The sum of all real value of \(x\) for which \(\frac{3x^2-9x+17}{x^2+3x+10}=\frac {5x^2-7x+19}{3x^2+5x+12}\) is equal to________.
The sum of all real value of \(x\) for which \(\frac{3x^2-9x+17}{x^2+3x+10}=\frac {5x^2-7x+19}{3x^2+5x+12}\) is equal to________.
Updated On: Jun 3, 2025
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Solution and Explanation
Given,
\(\frac{3x^2-9x+17}{x^2+3x+10}=\frac {5x^2-7x+19}{3x^2+5x+12}\)
⇒ \(\frac{x^2+3x+10+2x^2-12x+7}{x^2+3x+10}=\frac {3x^2+5x+12+2x^2-12x+7}{3x^2+5x+12}\)
⇒ \(1+\frac {2x^2-12x+7}{x^2+3x+10}=1+\frac {2x^2-12x+7}{3x^2+5x+12}\)
⇒ \(2x^2-12x+7 \frac{1}{x^2+3x+10}-\frac {1}{3x^2+5x+12}=0 \)
\(2x^2-12x+7=0\) Or \(3x^2+5x+12=x^2+3x+10\)
For \(2x^2-12x+7=0\)
Sum of roots=\(\frac {12}{2}=6 \)
Or \(3x^2+5x+12=x^2+3x+10\)
⇒ \(2x^2+2x+2=0 \)
⇒ \(x^2+x+1=0\)
No real roots as \(D=1-4<0\)
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Concepts Used:
Quadratic Equations
A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers.
Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.
The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)
Two important points to keep in mind are:
- A polynomial equation has at least one root.
- A polynomial equation of degree ‘n’ has ‘n’ roots.
Read More: Nature of Roots of Quadratic Equation
There are basically four methods of solving quadratic equations. They are:
- Factoring
- Completing the square
- Using Quadratic Formula
- Taking the square root