x=11+69/2 and x=11-69/2
-2.5 + 1.6583123951777i-2.5 - 1.6583123951777i
There are none because the discriminant of the given quadratic expression is less than zero.
To find the roots of the polynomial (x^2 + 3x - 5), we need to set the polynomial equal to zero and solve for x. So, (x^2 + 3x - 5 = 0). To solve this quadratic equation, we can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where a = 1, b = 3, and c = -5. Plugging these values into the formula, we get (x = \frac{-3 \pm \sqrt{3^2 - 41(-5)}}{2*1}), which simplifies to (x = \frac{-3 \pm \sqrt{29}}{2}). Therefore, the two values of x that are roots of the polynomial are (x = \frac{-3 + \sqrt{29}}{2}) and (x = \frac{-3 - \sqrt{29}}{2}).
A fifth degree polynomial.
Using the discriminant the possible values of k are -9 or 9
-2.5 + 1.6583123951777i-2.5 - 1.6583123951777i
There are none because the discriminant of the given quadratic expression is less than zero.
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
x = -2.5 + 1.6583123951777ix = -2.5 - 1.6583123951777iwhere i is the square root of negative one.
1
-2 and -6
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
None, it involves the square root of a negative number so the roots are imaginary.
-6 Check: -6+4-6+8 = 0
It is: 16-20 = -4 which means that the given quadratic expression has no real roots.
To find the roots of the polynomial (x^2 + 3x - 5), we need to set the polynomial equal to zero and solve for x. So, (x^2 + 3x - 5 = 0). To solve this quadratic equation, we can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where a = 1, b = 3, and c = -5. Plugging these values into the formula, we get (x = \frac{-3 \pm \sqrt{3^2 - 41(-5)}}{2*1}), which simplifies to (x = \frac{-3 \pm \sqrt{29}}{2}). Therefore, the two values of x that are roots of the polynomial are (x = \frac{-3 + \sqrt{29}}{2}) and (x = \frac{-3 - \sqrt{29}}{2}).
A fifth degree polynomial.