Solve the following quadratic equation: `\ x^2-(2+i)x-(1-7i)=0`

Options

A. `(3-i) and (-1+2i)`
B. `(3+i) and (-1+2i)`
C. `(3+i) and (-1-2i)`
D. `(3+i) and (1+2i)`

Answer

`A`

Explanation

`x^(2)-(2+i)x- (1-7i) =0`
Here, `a=1, b=- (2+ i), c=-(1- 7i)`
Now, from Sridharacharya's formula
` " "x=(-bpmsqrt(b^(2)-4ac))/(2a),`
`" "x=((2+ i)pmsqrt((2+ i)^(2)+ 4(1)( 1-7i)))/(2)`
`" "=((2+i)pmsqrt(4+i^(2)+4i+4- 28i))/(2)`
`rArr" "x=((2+i)pmsqrt(7-24i))/(2)" "...(1)`
Let `sqrt(7-24i)=p-iq`
`rArr" "7-24i=(p-iq)^(2)=p^(2)+i^(2)q^(2)-2ipq`
`" "=p^(2)-q^(2)-2ipq `
`therefore" "p^(2)-q^(2)=7" "...(2)`
and `" "2pq=24" "...(3)`
Now, `" "(p^(2)+q^(2))^(2)=( p^(2)-q^(2))^(2)+(2pq)^(2)=7^(2)+24^(2)`
`" "=625`
`rArr" "p^(2)+q^(2)=25" "...(4)`
From eqs. (2) and (4)
`" "2p^(2)=32`
`" "p^(2)=16`
`" "p=4`
From eq. (3)
`" "2(4)q=24`
`" "q=3`
`therefore" "sqrt(7-24i)=(4-3i)`
From eq. (1)
`x=((2+i)pm(4-3i))/(2)`
`" "=((2+i)+(4-3i))/(2)" or "((2+i)-(4-3i))/(2)`
`" "=(6-2i)/(2)" or " (-2+4i)/(2)`
`=(3-i)" or " (-1+2i)`
Therefore, the roots of given equation are (3-`i`) and `(-1+2i)`.