if the root of the equation x^2-10x+26=0 is 5+i, what is the other root?

Answer 1

Roots always occur in conjugate pairs. So that would be (5-i)
To see why this is, notice that all coefficients are real numbers, in fact, they are whole numbers. So you would need to multiply a pair of conjugates to get the sum of two square real numbers which results in another real number.
Also consider the quadratic formula, which generates a solution of the form (a +/- b). Graphically this translates to a parabola crossing the x axis at points b distance on either side of the vertex a. If the discriminant includes a complex number, you'd get a solution of the form (a +/- bi)

Answer 2

Using the quadratic formula, we get:
x = (10 +/- sqrt(100-4*26))/2 = 5 +/- i

Plugging in to check: (5-i)^2 - 10*(5-i)+26 = 25-10i-1-50+10i+26 = 0.

Answer 3

It will be (5 -i). This number is a complex conjugate of the (5 + i). When a complex number is multiplied by it's conjugate, the answer is real. A complex conjugate can be found by changing the sign of the imaginary part of the number, so the conjugate of a - bi is a + bi. Any quadratic that has real coefficients and non-real roots will have a pair of conjugates as the roots. This can be see from the quadratic formula.

Answer 4

(x - (5 + i))(x - (5 - i))
x^2 - (5 + i)x - (5 - i)x + ((5 + i)(5 - i))
x^2 - 5x - ix - 5x + ix + (25 - i^2)
x^2 - 10x + (25 - (-1))
x^2 - 10x + (25 + 1)
x^2 - 10x + 26

ANS : 5 - i

Answer 5

-5-i

Answer 6

for
ax² + bx +c =0
x = (-b ±√(b² - 4ac) )/2a
x = 10/2 ± √-4 /4
x = 5 + i OR 5 - i

Answer 7

Use the quadratic formula.

Answer 8

why are you doing math over x-mas break?