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Quadratic Equation With Real Coefficients Imaginary, Irrational Roots

Quadratic equation with real coefficients Imaginary, 
                              Irrational Roots occur in conjugate pairs Online Tutoring / Homework Help

THEOREM:


In a quadratic equation with real coefficients imaginary roots occur in conjugate pairs.

Proof:

Let us consider the quadratic equation in the general form:

ax2 + bx + c = 0               ****** (1)

In this equation the coefficients a, b and c are real.

Let the imaginary quantity Alpha + i [Alpha, are real and i =√-1] be a root of equation (1).

Therefore, it should be satisfied with the equation x = Alpha + i.

Then, a (Alpha + i)2 + b (Alpha + i) + c = 0

Or, a(Alpha2 - 2 + i • 2Alpha) + bAlpha + ib + c = 0 [As, i2 = -1]

Or, aAlpha 2 - a2 + bAlpha + c + i(2aAlpha + b) = 0

Since, aAlpha2 - a2 + bAlpha + c = 0         ******(2)

and 2aAlpha + b = 0         ******(3)

[Since p +iq = 0 (p, q are real and i = √-1) means p = 0 and q = 0]

Therefore, replacing x in ax2 + bx + c by Alpha - i we get,

a(Alpha - i)2 + b (Alpha - i) + c

= a (Alpha 2 - 2 - i 2Alpha) + bAlpha - ib + c

= aAlpha 2 - Alpha 2 + bAlpha + c - i (2aAlpha + b) = 0 - i • 0      [from (2) and (3)] = 0

This shows that the equation (1) is satisfied by x = Alpha - i when (Alpha + i) is a root of the equation. Then, (Alpha - i) is the other root of the equation. Similarly, if (Alpha - i) is a root of equation (1), it can be proved that its other root is (Alpha + i). Again, (Alpha + i) and ( Alpha - i) are conjugate complex quantities. So, we conclude that in a quadratic equation imaginary roots occur only in conjugate pairs.


THEOREM:



In a quadratic equation with rational coefficients irrational roots occur in conjugate pairs.

Proof:

Let us consider the quadratic equation in the general form:

ax2 + bx + c = 0            ******(1)

In this equation the coefficients a, b and c are rational.

Let p + √q (In this equation, p is rational and √q is irrational) be a root of equation (1).

Then the equation (1) should be satisfied with x = p + √q.

Therefore, a(p + √q)2 + b( p + √q) + c = 0

Or,              a(p2 + q + 2p√q) + bp + b√q + c = 0

Or,               ap2 + aq + bp + c + (2ap + b ) √q = 0 = 0 + 0. √q

Therefore, ap2 + aq + bp + c = 0              ******(2)      and 2ap + b = 0             ******(3)

Therefore, replacing x in ax2 + bx + c by p - √q we get,

a(p - √q)2 + b( p - √q) + c

= a(p2 + q - 2p√q) + bp - b√q + c

= ap2 + aq + bp + c - ( 2ap + b ) √q = 0 -√q.0              [ by using equation (2) and (3) ] = 0

This shows that the equation (1) is satisfied by x = p - √q when ( p + √q) is a root of the equation. Then, ( p - √q) is the other root of the equation. Similarly, if (p - √q) is a root of equation (1), it can be proved that its other root is ( p + √q). Again, ( p + √q) and ( p - √q) are conjugate irrational quantities. So, we conclude that in a quadratic equation irrational roots occur only in conjugate pairs.


Example 1:

Find the quadratic equation with real coefficients which has 2 + i as a root (i = √-1)

Solution:

By the question, coefficients of the required quadratic equation are real and its one root is 2 + i. Hence, the other root of the required equation is 2 - i [Since, in a quadratic equation with real coefficients imaginary roots occur in conjugate pairs].


Now, the sum of the roots of the equation = 2 + i +2 - i = 4.

And the product of the roots = (2 + i)(2 - i) = 4 - i2

                                                 = 4 + 1 [Since, i =√-1 and since,i2 = -1] = 5.

Therefore, the required equation is x2 - 4x + 5 = 0.       (Answer)

Example 2:

Find a quadratic equation with rational coefficient whose one root is 3 - √5.

Solution:

By the question, coefficients of the required quadratic equation are rational and its one root is 3 - √5. Hence, the other root of the required equation is 3 + √5 [Since, in a quadratic equation with rational coefficients irrational roots occur in conjugate pairs].

Therefore, the required equation is

x2 - (sum of the roots)x + product of the roots = 0

Or, x2 - (3 - √5 + 3 + √5)x + (3 - √5)(3 + √5) = 0

Or, x2 - 6x + 9 - 5 = 0

Or, x2 - 6x + 4 = 0.         (Answer)

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