Solve this quadratic equation using the quadratic formula. 3×2 + 5x + 1 = 0
The Correct Answer and Explanation is:
To solve the quadratic equation 3×2+5x+1=03x^2 + 5x + 1 = 03×2+5x+1=0 using the quadratic formula, we will first recall the general quadratic formula:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac
Here, the coefficients from the equation 3×2+5x+1=03x^2 + 5x + 1 = 03×2+5x+1=0 are:
- a=3a = 3a=3
- b=5b = 5b=5
- c=1c = 1c=1
Step 1: Identify the coefficients
From the given quadratic equation:
- a=3a = 3a=3
- b=5b = 5b=5
- c=1c = 1c=1
Step 2: Substitute into the quadratic formula
Substitute the values of aaa, bbb, and ccc into the quadratic formula:x=−5±52−4(3)(1)2(3)x = \frac{-5 \pm \sqrt{5^2 – 4(3)(1)}}{2(3)}x=2(3)−5±52−4(3)(1)
Step 3: Simplify the discriminant
First, calculate the discriminant (the part under the square root):b2−4ac=52−4(3)(1)=25−12=13b^2 – 4ac = 5^2 – 4(3)(1) = 25 – 12 = 13b2−4ac=52−4(3)(1)=25−12=13
Now, substitute the discriminant value back into the formula:x=−5±136x = \frac{-5 \pm \sqrt{13}}{6}x=6−5±13
Step 4: Solve for xxx
We now have two possible solutions for xxx because of the ±\pm± sign:x=−5+136orx=−5−136x = \frac{-5 + \sqrt{13}}{6} \quad \text{or} \quad x = \frac{-5 – \sqrt{13}}{6}x=6−5+13orx=6−5−13
Step 5: Approximate the solutions
For practical purposes, we can approximate the square root of 13:13≈3.6055\sqrt{13} \approx 3.605513≈3.6055
Now, substitute this value into the two solutions:
- x=−5+3.60556=−1.39456≈−0.2324x = \frac{-5 + 3.6055}{6} = \frac{-1.3945}{6} \approx -0.2324x=6−5+3.6055=6−1.3945≈−0.2324
- x=−5−3.60556=−8.60556≈−1.4342x = \frac{-5 – 3.6055}{6} = \frac{-8.6055}{6} \approx -1.4342x=6−5−3.6055=6−8.6055≈−1.4342
Final Solutions:
The solutions to the quadratic equation 3×2+5x+1=03x^2 + 5x + 1 = 03×2+5x+1=0 are approximately:x≈−0.2324orx≈−1.4342x \approx -0.2324 \quad \text{or} \quad x \approx -1.4342x≈−0.2324orx≈−1.4342
These are the two real roots of the equation.
