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

Solve for z

z=\frac{-1+3\sqrt{11}i}{2}\approx -0.5+4.974937186i

z=\frac{-3\sqrt{11}i-1}{2}\approx -0.5-4.974937186i

Solve z^2+z+25=0 | Microsoft Math Solver (1)

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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.

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

FAQs

What is the number of solutions of the equation Z2 z 2 0? ›

Hence, Z2 + |Z|2 = 0 has infinity many solutions.

What is the website that solves any math problem? ›

Mathway. Mathway calculator is a smart math problem solver which gives you a step by step solution to a math problem. Just type your question and press enter to reveal a detailed answer to your math problem. This intelligent math problem solver, simplifies the process with step by step solutions.

Does z 2 =| z 2? ›

By definition, |z|2=z¯z, | z | 2 = z z ¯ , that is, the product of z∈C z ∈ C by its conjugate. hence b=0. So we have that |z|2=z2 | z | 2 = z 2 if and only if z is real.

What are the solutions to the quadratic equation x2 + 25 = 0 brainly? ›

Final answer:

The solutions to the quadratic equation x²−25=0 are x=5 and x=−5. These are obtained by factoring the difference of squares into (x+5)(x-5)=0, which leads to two real solutions.

Which equation has 2 solutions? ›

A quadratic equation with real or complex coefficients has two solutions, called roots.

How many solutions are there for the equation z 2 conjugate of z? ›

Hence, the no. of solution of equationz2=¯¯¯z is 4 solution.

Is Microsoft Math Solver free? ›

All for free and now with a games section! HOW IT WORKS: Use your camera to scan the problem, write it out on our whiteboard, or enter it in our calculator.

How to check math answers? ›

Plug the solution back into the equation.

This is the simplest way to check that your answer is correct. If you solved for a variable or multiple variables, plug these solutions back into the equation and work backwards to see if they make the equation true. If they do, then the solutions are correct.

What does z mean in Algebra 2? ›

The letter (Z) is the symbol used to represent integers. An integer can be 0, a positive number to infinity, or a negative number to negative infinity. The integers (Z): . . . -3, -2, -1, 0, 1, 2, 3 . . . Notice that every whole number is an integer.

How to find root i? ›

Find the square root of i
  1. i = exp(iπ/2) Now the expression is in exponential form, taking the square root is easy, using basic exponential math.
  2. sqrt(i) = (exp(iπ/2))^(1/2) = exp(iπ/4) This quantity has a modulus of 1 and an argument of π/4. Using Euler's formula again,
  3. sqrt(i) = (1 + i)/sqrt(2)

When the number z 2 z 2 is purely imaginary? ›

Then z−2z+2 is purely imaginary implies that the angle between the vector joining z and −2, and the vector joining z and 2, is π2, as argument of any imaginary number is π2. Now you can see that, this condition implies z lies on the circumfurence of the circle with diameter whose endpoints are 2 and −2.

How many real roots does the quadratic equation x2 25 0 have? ›

The roots of given quadratic equation is ±5. This is a second order equation that means it will have two values of x.

How can you tell that a quadratic equation is going to have no solution? ›

Answer and Explanation:

(2) If a quadratic equation has a negative discriminant then there will be no real solution of the equation. So equation x 2 + x + 1 = 0 does not contain any real solution.

What value of b satisfy 3 2b 3 2 36? ›

So, the required values of b are (2√3 – 3)/2 and (-2√3 + 3)/2.

How do you find the number of solutions in an equation? ›

If we can solve the equation and get something like x=b where b is a specific number, then we have one solution. If we end up with a statement that's always false, like 3=5, then there's no solution. If we end up with a statement that's always true, like 5=5, then there are infinite solutions..

What are the solutions of the equation z2 12z 36 0? ›

Summary: The solution of the equation: z2 - 12z + 36 is z = 6.

How many solutions can be found for the equation 4z 2 z − 4 3z 11? ›

Answer and Explanation:

There is only one solution for the equation 4z + 2(z -4) = 3z + 11 because the exponent for the power of z is 1.

How many solutions does the following equation have − 5 z 1 − 2 z 10 − 5 z 1 )= − 2z 10? ›

If − 5 ( z + 1 ) = − 2 z + 10 then there is one valid solution, z = -5.

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