1)Solve and write in standard form:
4x^2-4x+21=0
2)Find the standard form of the complex number.
8(cos pi/2 + i sin pi/2)
Thanks everyone!
x = (4 ± √(16 - 16(21)) )/8
= (4 ± √-320)/8
= (4 ± 8√-5)/8
= 1/2 ± i√5
x = 1/2 + i√5 or x = 1/2 - i√5
2)
8(cos π/2 + i sin π/2)
= 8(√2/2 + i √2/2)
= 4√2 + i (4√2)
To solve the quadratic equation in the first question, we can use the quadratic formula. The standard form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are coefficients.
Given the equation 4x^2 - 4x + 21 = 0, we can identify that a = 4, b = -4, and c = 21.
The quadratic formula states that for an equation ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Let's substitute the values of a, b, and c into the formula.
x = (-(-4) ± √((-4)^2 - 4 * 4 * 21)) / (2 * 4)
x = (4 ± √(16 - 336)) / 8
x = (4 ± √(-320)) / 8
Here, we have a complex number with a negative under the square root. To simplify further, we can write -320 as -1 * 32 * 10. Then we have:
x = (4 ± 4i√(10)) / 8
Further simplifying:
x = (1 ± i√(10)) / 2
Therefore, the solutions to the quadratic equation 4x^2 - 4x + 21 = 0, when written in standard form, are:
x = (1 + i√(10)) / 2 and x = (1 - i√(10)) / 2.
Moving on to the second question, we are given a complex number in exponential form, 8(cos(π/2) + i sin(π/2)).
The standard form of a complex number is a + bi, where a and b are real numbers.
To convert the complex number from exponential form to standard form, we can use Euler's formula:
e^(ix) = cos(x) + i sin(x)
Here, we have 8(cos(π/2) + i sin(π/2)), which matches the form e^(ix).
By applying Euler's formula, we can rewrite the complex number as:
8e^(iπ/2)
Using the exponential form, we can convert it to the standard form:
8(cos(π/2) + i sin(π/2)) = 8(0 + i * 1) = 8i
Therefore, the standard form of the complex number 8(cos(π/2) + i sin(π/2)) is 8i.