If 2 + i and √5-2i are the roots of the equation (x2+ax+b)(x2+cx+d)=0, where a, b, c, d are real constants, then product of all roots of the equation is

Question

If 2 + i and √5-2i are the roots of the equation (x2+ax+b)(x2+cx+d)=0, where a, b, c, d are real constants, then product of all roots of the equation is (A) 40 (B) 9√5 (C) 45 (D) 35

Tardigrade Admin · Accepted Answer

If one root of a quadratic equation is of the form

a + ib

, then other root will be

a - ib

.
So, all the roots are

2 \pm i, 5 \pm 2 i

∴

Product of all the roots

= (2 + i) (2 - i) (5 + 2 i) (5 - 2 i)

= (4 + 1) (5 + 4) = 5 \times 9 = 45

Q. If $2 + i$ and $5 - 2 i$ are the roots of the equation $(x^{2} + a x + b) (x^{2} + c x + d) = 0,$ where $a, b, c, d$ are real constants, then product of all roots of the equation is

$40$

$95$

$45$

$35$

If one root of a quadratic equation is of the form $a + ib$ , then other root will be $a - ib$ .
So, all the roots are $2 \pm i, 5 \pm 2 i$
$∴$ Product of all the roots
$= (2 + i) (2 - i) (5 + 2 i) (5 - 2 i)$
$= (4 + 1) (5 + 4) = 5 \times 9 = 45$

Q. If 2+i and 5​−2i are the roots of the equation (x2+ax+b)(x2+cx+d)=0, where a,b,c,d are real constants, then product of all roots of the equation is

40

95​

45

35