Question

x²-15x+8=0 quadratic equations​

Answer 1

Answer:

Either x = $\frac{15+\sqrt{193}}{2}$ or $\frac{15-\sqrt{193}}{2}$.

Step-by-step explanation:

$\sqrt{15^2-(4)(1)(8)}=\sqrt{193}>0$

Therefore, the roots of the quadratic equation are distinct real numbers. (As the the discriminant > 0.)

The Quadratic formula is: x = $-b$±$\sqrt{b^2-4ac}$/$2a$

Comparing with the standard form of Quadratic Equations, $ax^{2}+bx+c=0$, a = 1, b = -15, c = 8.

Using the Quadratic formula, x = -(-15)±$\sqrt{(-15)^2-(4)(1)(8)}$/2(1) = 15±$\sqrt{193}$/2.

So, either x = $\frac{15+\sqrt{193} }{2}$ or $\frac{15-\sqrt{193} }{2}$.