-3x^2 + 2x – 5 = 0
The Correct Answer and Explanation is:
To solve the quadratic equation −3×2+2x−5=0-3x^2 + 2x – 5 = 0−3×2+2x−5=0, we can use the quadratic formula, which is given by:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac
For the equation −3×2+2x−5=0-3x^2 + 2x – 5 = 0−3×2+2x−5=0, we have:
- a=−3a = -3a=−3
- b=2b = 2b=2
- c=−5c = -5c=−5
Now, substitute these values into the quadratic formula:x=−2±(2)2−4(−3)(−5)2(−3)x = \frac{-2 \pm \sqrt{(2)^2 – 4(-3)(-5)}}{2(-3)}x=2(−3)−2±(2)2−4(−3)(−5)
First, calculate the discriminant (b2−4ac)(b^2 – 4ac)(b2−4ac):b2−4ac=(2)2−4(−3)(−5)=4−60=−56b^2 – 4ac = (2)^2 – 4(-3)(-5) = 4 – 60 = -56b2−4ac=(2)2−4(−3)(−5)=4−60=−56
Since the discriminant is negative (−56-56−56), it means that the equation has complex solutions. Now, proceed with the solution:x=−2±−56−6x = \frac{-2 \pm \sqrt{-56}}{-6}x=−6−2±−56
The square root of −56-56−56 is −56=56i=214i \sqrt{-56} = \sqrt{56}i = 2\sqrt{14}i−56=56i=214i, where iii is the imaginary unit. Now, substitute this back into the equation:x=−2±214i−6x = \frac{-2 \pm 2\sqrt{14}i}{-6}x=−6−2±214i
Simplify the expression by dividing both terms by −6-6−6:x=13∓14i3x = \frac{1}{3} \mp \frac{\sqrt{14}i}{3}x=31∓314i
Thus, the two complex solutions are:x=13+14i3orx=13−14i3x = \frac{1}{3} + \frac{\sqrt{14}i}{3} \quad \text{or} \quad x = \frac{1}{3} – \frac{\sqrt{14}i}{3}x=31+314iorx=31−314i
Explanation:
This equation is quadratic, meaning its solutions can either be real or complex. In this case, the discriminant (b2−4ac)(b^2 – 4ac)(b2−4ac) is negative, indicating complex solutions. When you encounter a negative discriminant, the solutions will always involve imaginary numbers.
