Question

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Answer 1

Answer:

Step-by-step explanation:

Here, let a=1,b=-(7-i) and c=(18-i)

Applying the quadratic formula,

x=(-b±√(b^2-4ac))/2a

x=(-(-(7-i))±√((-(7-i))^2-4(1)(19-i)))/2(1)

=(7-i±√(49-14i+i^2-76+4i))/2

=(7-i±√(i^2-10i-25))/2

=(7-i±√((i-5)^2))/2

=(7-i±(i-5))/2

Therefore, x=(7-i+(i-5))/2,(7-i-(i-5))/2

=2/2,(12-2i)/2

x=1,(6-i)

Answer 2

Step-by-step explanation:

we have given equation ,

$\sf x^2 - (7 - i)x + (18 - i) = 0$

first of all, check whether the roots are, real or not

compare the given equation with

$\sf ax^2 + bx + c$

a = 1 , b = -(7 - i) , c = (18 - i)

$\sf Delta = b^2 - 4ac$

$\sf= (-(7-i)) ^2 - 4(1)(18-i)$

$\sf= (7-i)^2 -72 + 4i$

$\sf = (7-i)^2 - 72 + 4i$

$\sf= 49 - 14i + i^2 - 72 + 4i$

$\sf = -23 - 10i - 1$

$\sf= -10i - 24$

therefore the equation has complex roots

so,

$\sf x = \frac{-b \pm \sqrt{ b^2 - 4ac}}{2a}$

$\sf x = \frac{-(-(7-i) \pm \sqrt {(-(7-i))^2 - 4(1)(18 - i) }}{2 (1)}$

$\sf x = \frac {(7 - i) \pm \sqrt { (7 - i)^2 - 72 + 4i }}{2}$

$\sf x = \frac { (7 - i) \pm \sqrt { 49 -14 i + i^2 - 72 + 4i }}{2}$

$\sf x = \frac{(7 - i) \pm \sqrt { -23 - 14i - 1 + 4i }}{2}$

$\sf x = \frac{ (7 - i) \pm \sqrt {-10i - 24}}{2}$

Now, consider

$\sf \ sqrt {-10i - 24} = p + qi$ , here p, q € R

squaring on both the sides,

$\sf (p + qi) ^2 = -10i - 24$

$\sf (p^2 + 2p (qi) + (qi) ^2 = -10i - 24$

$\sf (p^2 + 2pqi - q^2 ) =-10i - 24$

$\sf (p^2 - q^2 ) + 2pqi = - 24- 10i$

Now, equate the real and imaginary values,

$\sf p^2 - q^2 = - 24$ ----(1)

$\sf 2pq = - 10$

$\sf q = \frac{-10}{2p}$

$\sf q = \frac{-5}{p}$

put the value of q in equation 1

$\sf p^2 - (\frac{-5}{p})^2 = - 24$

$\sf p^2 - (\frac{25}{p^2}) = -24$

$\sf (\frac{p^4 - 25}{p^2})= -24$

$\sf p^4 - 25 = -24p^2$

$\sf p^4 + 24p^2 - 25$

on solving we get ,

$\sf (p^2 - 1)(p^2 + 25 )$

$\sf p^2 = -25$

here, $\sf p^2 ≠ - 25$ ...........(not real number)

therefore,

$\sf p = \pm 1$

by putting the value of p in

$q = \frac{-10}{2p}$

we get , $q = \pm 5$

here the signs are matters, when the sign of p is negative then the sign of q will be positive and when sign of p is positive then sign of q is negative.

therefore ,$\sf \sqrt {-10i - 24} = \pm (1 - 5i)$

Now,

$\sf x = \frac{ (7 - i) \pm \sqrt {-10i - 24}}{2}$

$\sf x = \frac{(7-i)\pm (1 - 5i) }{2}$

therefore the values of x are,

$\longrightarrow\sf x = \frac { 7 - i + 1 - 5i }{2}$

$\sf x = \frac{8 -4i}{2}$

$\sf \boxed{x =4 - 2i }$

OR

$\longrightarrow\sf x = \frac { 7 - i -( 1 - 5i) }{2}$

$\longrightarrow\sf x = \frac{7 - i - 1 + 5i }{2}$

$\longrightarrow\sf x =\frac{6 + 4i }{2}$

$\sf\boxed{ x = 3 + 2i }$