Find the values of A,B,C and D if 1+2i and 1+3i are the roots of the equation x^5-7x^4+Ax^3+Bx^2+Cx+D=0?

Answer 1

x^5 - 7x^4 + Ax^3 +Bx^2 + Cx + D = 0

Two things to note:
1) By the Fundamental Theorem of Algebra, every equation of degree n has exactly n roots (including multiplicity and complex roots.)
2) Complex roots ALWAYS come in conjugate pairs.

So what does this mean? This means that since we know (1 + 2i) and (1 + 3i) are roots, it follows that their conjugate counterparts, (1 - 2i) and (1 - 3i), are also roots. That's 4 roots right there; we can easily find the 5th if we wanted.

Since (1 + 2i) is a root, (x - (1 + 2i)), is a factor.
Since (1 - 2i) is a root, (x - (1 - 2i)) is a factor.
Since (1 + 3i) is a root, (x - (1 + 3i)) is a factor.
Since (1 - 3i) is a root, (x - (1 - 3i)) is a factor.

Now, we multiply these all out. Let's multiply the first two factors first together, and then the last two factors. What we're going to do is then multiply these two together.

[x - (1 + 2i)] [x - (1 - 2i)] = x^2 - x(1 - 2i) - x(1 + 2i) + (1 + 2i)(1 - 2i)

Doing some expanding and simplification, we get

= x^2 - x + (2i)x - x - (2i)x + 1 - 4i^2

Remember that i^2 = -1, so

= x^2 - 2x + 1 - 4(-1)
= x^2 - 2x + 1 + 4
= x^2 - 2x + 5

[x - (1 + 3i)] [x - (1 - 3i)] = x^2 - x(1 - 3i) - x(1 + 3i) + (1 + 3i)(1 - 3i)

Expanding this, we have

= x^2 - x + (3i)x - x - (3i)x + (1 - 9i^2)
= x^2 - 2x + 1 - 9(-1)
= x^2 - 2x + 10

Now, we multiply (x^2 - 2x + 5) and (x^2 - 2x + 10) together.

(x^2 - 2x + 5)(x^2 - 2x + 10) =
x^4 - 2x^3 + 10x^2 - 2x^3 + 4x^2 - 20x + 5x^2 - 10x + 50 =
x^4 - 4x^3 + 19x^2 - 30x + 50

At this point, we *have* 4 roots and need the 5th one. We know the 5th root HAS TO BE A REAL NUMBER, because complex roots always come in conjugate pairs Therefore, it will follow that (x - r) is a factor. Just multiply (x - r) with the stuff above.

(x - r) (x^4 - 4x^3 + 19x^2 - 30x + 50) =
x^5 - 4x^4 + 19x^3 - 30x^2 + 50 - rx^4 + 4rx^3 - 19rx^2 + 30rx - 50r

Combining like terms,

x^5 + (-4 - r)x^4 + (19 + 4r)x^3 + (-30 - 19r)x^2 + (30r)x + (-50r)

Now, we can equate each coefficient componentwise with the original equation. Note that the coefficient of x^4 in the original equation is -7, and in this new equation, -4 - r. Therefore,

-7 = -4 - r, so
r = -4 + 7
r = 3

So we now know the value of r, and our equation is

x^5 + (-4 - 3)x^4 + (19 + 4(3))x^3 + (-30 - 19(3))x^2 + (30(3))x + (-50(3)) =

x^5 - 7x^4 + 31x^3 - 87x^2 + 90x - 150

And as you can see,
A = 31
B = -87
C = 90
D = -150

Answer 2

x^5-7x^4+Ax^3+Bx^2+Cx+D=0?
If 1 + 2i and 1 +3i are its roots, then 1 - 2i and 1 - 3i are also its roots because complex roots occur in pairs of complex conjugate numbers. Let the fifth root be a,
then the sum of the roots = 4 + a =7
which leads to a = 3. This should satisfy the given equation , so : 27 A + 9 B + 3 C +D = 324..............(1)
Product of the roots =(1+2i) (1-2i)(1+3i)(1-3i) 3 = -D
or 5*10* 3 = - D or D = - 150 Answer
Sum of the products of roots taken two at a time
= 3(1+2i) + 3( 1-2i) +3(1+3i)+ 3 (1-3i) +(1+2i)(1-2i) +(1+3i)(1-3i)
+(1+2i)(1+3i) +(1+2i) (1-3i)+ (1-2i)(1+3i) + (1-2i)(1-3i) = A
So : A = 3*4 +5 +10 +2(1+2i) +2(1-2i) = 31. Or A =31Answer
Similarly you can find B and C, by taking sum of products of roots taken 3 at a time and 4 at a time..

Answer 3

Complex roots come in conjugates, so it is
(x - 1 - 2i)(x - 1 + 2i)(x - 1 - 3i)(x - 1 + 3i)(x - a)

And, the sum of the roots is the negative of the n - 1 term (in this case, n = 5, so it is the x^4 term)

1 + 2i + 1 - 2i + 1 + 3i + 1 - 3i + a = 7, so a = 3
(x^2 - 2x + 1 + 4)(x^2 - 2x + 1 + 9)(x - 3)
(x^2 - 2x + 5)(x^2 - 2x + 10)(x - 3)

You can multiply this out to get the answer.

Answer 4

x^5 - 7x^4 + Ax^3 +Bx^2 + Cx + D = 0

Two things to note:
1) By the Fundamental Theorem of Algebra, every equation of degree n has exactly n roots (including multiplicity and complex roots.)
2) Complex roots ALWAYS come in conjugate pairs.

So what does this mean? This means that since we know (1 + 2i) and (1 + 3i) are roots, it follows that their conjugate counterparts, (1 - 2i) and (1 - 3i), are also roots. That's 4 roots right there; we can easily find the 5th if we wanted.

Since (1 + 2i) is a root, (x - (1 + 2i)), is a factor.
Since (1 - 2i) is a root, (x - (1 - 2i)) is a factor.
Since (1 + 3i) is a root, (x - (1 + 3i)) is a factor.
Since (1 - 3i) is a root, (x - (1 - 3i)) is a factor.

Now, we multiply these all out. Let's multiply it out two at a time.

[x - (1 + 2i)] [x - (1 - 2i)] = x^2 - x(1 - 2i) + x(1 + 2i) + (1 + 2i)(1 - 2i)
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Answer 5

advance all of the brackets: (6 + 5i - 6i^2) / (2 + 3i - 2i^2) undergo in innovations that i^2 = -one million: (12 + 5i) / (4 + 3i) Now rationalise the denominator by skill of multiplying the two the numerator and denominator by skill of the denominator yet with different sign (4 - 3i). be conscious how the two imaginary numbers cancel one yet yet another out: (12 + 5i) (4 - 3i) = 40 8 - 36i + 20i - 15i^2 = 40 8 - 16i - 15i^2 = sixty 3 - 16i (4 + 3i) (4 - 3i) = sixteen - 12i + 12i - 9i^2 = sixteen - 9i^2 = 25 As a fraction: (sixty 3 - 16i) / 25

Answer 6

To add on to Puggy's note:

Multiply each of the conjugate pairs together (1+2i)(1-2i)

(a+bi)(a-bi) = a²+b²

You get a real number
You also know that [x² - (a²+b²)] is the product of two roots to the equation. Do that twice and you should have all four roots to the equation. Multiply that all together and you should have it all.

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edit. Oops.. I missed the 5th power on re-reading it. Yeah, dividing it out should give you the last root.

Answer 7

well, hell if i kno...i aint no scientist...lol