Solve z^2+z+25=0 | Microsoft Math Solver (2024)

https://math.stackexchange.com/questions/1583128/for-what-r-s-exist-unbiased-estimation-of-fp-pr1-ps-for-binomi

z^{2}+z+25=0

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

z=\frac{-1±\sqrt{1^{2}-4\times 25}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

z=\frac{-1±\sqrt{1-4\times 25}}{2}

Square 1.

z=\frac{-1±\sqrt{1-100}}{2}

Multiply -4 times 25.

z=\frac{-1±\sqrt{-99}}{2}

Add 1 to -100.

z=\frac{-1±3\sqrt{11}i}{2}

Take the square root of -99.

z=\frac{-1+3\sqrt{11}i}{2}

Now solve the equation z=\frac{-1±3\sqrt{11}i}{2} when ± is plus. Add -1 to 3i\sqrt{11}.

z=\frac{-3\sqrt{11}i-1}{2}

Now solve the equation z=\frac{-1±3\sqrt{11}i}{2} when ± is minus. Subtract 3i\sqrt{11} from -1.

z=\frac{-1+3\sqrt{11}i}{2} z=\frac{-3\sqrt{11}i-1}{2}

The equation is now solved.

z^{2}+z+25=0

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

z^{2}+z+25-25=-25

Subtract 25 from both sides of the equation.

z^{2}+z=-25

Subtracting 25 from itself leaves 0.

z^{2}+z+\left(\frac{1}{2}\right)^{2}=-25+\left(\frac{1}{2}\right)^{2}

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

z^{2}+z+\frac{1}{4}=-25+\frac{1}{4}

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

z^{2}+z+\frac{1}{4}=-\frac{99}{4}

Add -25 to \frac{1}{4}.

\left(z+\frac{1}{2}\right)^{2}=-\frac{99}{4}

Factor z^{2}+z+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

\sqrt{\left(z+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{99}{4}}

Take the square root of both sides of the equation.

z+\frac{1}{2}=\frac{3\sqrt{11}i}{2} z+\frac{1}{2}=-\frac{3\sqrt{11}i}{2}

Simplify.

z=\frac{-1+3\sqrt{11}i}{2} z=\frac{-3\sqrt{11}i-1}{2}

Subtract \frac{1}{2} from both sides of the equation.

x ^ 2 +1x +25 = 0

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

r + s = -1 rs = 25

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

r = -\frac{1}{2} - u s = -\frac{1}{2} + u

Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(-\frac{1}{2} - u) (-\frac{1}{2} + u) = 25

To solve for unknown quantity u, substitute these in the product equation rs = 25

\frac{1}{4} - u^2 = 25

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = 25-\frac{1}{4} = \frac{99}{4}

Simplify the expression by subtracting \frac{1}{4} on both sides

u^2 = -\frac{99}{4} u = \pm\sqrt{-\frac{99}{4}} = \pm \frac{\sqrt{99}}{2}i

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r =-\frac{1}{2} - \frac{\sqrt{99}}{2}i = -0.500 - 4.975i s = -\frac{1}{2} + \frac{\sqrt{99}}{2}i = -0.500 + 4.975i

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.