The roots of the equation are [5 +/- sqrt(11)]/2 = 4.158 and 0.842
2x2 + 3x - 20 = (x + 4)(2x - 5).
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.
No, an equation needs an "=".
It often helps to take square roots on both sides of the equation. However, solutions to the original equation may be lost - it is often convenient to put a "plus or minus" sign so as not to lose solutions. Example: x2 = 25 Taking square roots: x = "plus or minus" 5
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
The roots are: x = -5 and x = -9
There are no real root. The complex roots are: [-5 +/- sqrt(-3)] / 2
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
The roots of the equation are [5 +/- sqrt(11)]/2 = 4.158 and 0.842
It is a quadratic equation with one unknown variable, x which has no real roots.
2x2 + 3x - 20 = (x + 4)(2x - 5).
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.
No, an equation needs an "=".
It often helps to take square roots on both sides of the equation. However, solutions to the original equation may be lost - it is often convenient to put a "plus or minus" sign so as not to lose solutions. Example: x2 = 25 Taking square roots: x = "plus or minus" 5
Using the discriminant of b^2 -4ac = 0 the value of k works out as -2
The equation [ y = x2 - 2x + 5 ] has an infinite number of solutions,corresponding to every point on the graph of it.The ROOTS corresponding to the two values of 'x' that satisfy the equation when y=0 ...in other words, the points where the graph crosses the x-axis.In the particular case of this equation, the graph doesn't cross the x-axis at all,and the roots are not real numbers. There are still two of them, and they arecomplex conjugates.The roots areX = 1 ± 2 i