Question:

Solve the following quadratic equation by formula method: \[ x^2 + 10x + 2 = 0 \]

Show Hint

Use the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve any quadratic equation. The term \(b^2 - 4ac\) (discriminant) helps determine the nature of the roots.
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

Step 1: Identify the coefficients.
The given equation is \(x^2 + 10x + 2 = 0\). Comparing with \(ax^2 + bx + c = 0\), we get: \[ a = 1, \quad b = 10, \quad c = 2 \] Step 2: Recall the quadratic formula.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Step 3: Substitute the values.
\[ x = \frac{-10 \pm \sqrt{(10)^2 - 4(1)(2)}}{2(1)} \] \[ x = \frac{-10 \pm \sqrt{100 - 8}}{2} \] \[ x = \frac{-10 \pm \sqrt{92}}{2} \] Step 4: Simplify.
\[ x = \frac{-10 \pm 2\sqrt{23}}{2} = -5 \pm \sqrt{23} \] Step 5: Conclusion.
Hence, the roots of the quadratic equation are \(x = -5 + \sqrt{23}\) and \(x = -5 - \sqrt{23}.\)
Final Answer: \[ \boxed{x = -5 + \sqrt{23} \quad \text{and} \quad x = -5 - \sqrt{23}} \]
Was this answer helpful?
0
0

Top Questions on Quadratic Equations

View More Questions

Questions Asked in Maharashtra Class X Board exam

View More Questions