8
x2 - 2x - 15 = 0
x - 4y = 3 is the slope of a line perpendicular to line whose equation is y -5 3x plus 8 3.
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}).
If you mean y = 2x+5 then the perpendicular slope is -1/2
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
19
You don't have an equation there.
find the sum and product of the roots of 8×2+4×+5=0
x2 + 2x + 5 = 0 is already in the form ax2 + bx + c = 0 so to find the discriminant, D, you use D = b2 - 4ac Then if D is greater than 0 the equation has 2 real roots; if D = 0 the equation has one real root and if D is less than 0 the equation has no real roots. So to check this we work out D but we need to know what a, b and c are. From the equation we can see that a = 1 b = 2 c = 5 so putting these values in to find D: D = (2)2 - 4(1)(5) = 4 - 20 = -16 so the equation x2 + 2x + 5 = 0 has no real roots.
-3
8
(x - 3) and (x - (-5)).
The complex roots of an equation is any solution to that equation which cannot be expressed in terms of real numbers. For example, the equation 0 = x² + 5 does not have any solution in real numbers. But in complex numbers, it has solutions.
The roots are: x = -5 and x = -9
The roots of the equation are [5 +/- sqrt(11)]/2 = 4.158 and 0.842