Irrational Roots of a Quadratic Equation

We will discuss about the irrational roots of a quadratic equation.

In a quadratic equation with rational coefficients has a irrational or surd root α + √β, where α and β are rational and β is not a perfect square, then it has also a conjugate root α - √β.

Proof:

To prove the above theorem let us consider the quadratic equation of the general form:

ax\(^{2}\) + bx + c = 0 where, the coefficients a, b and c are real.

Let p + √q (where p is rational and √q is irrational) be a surd root of equation ax\(^{2}\) + bx + c = 0. Then the equation ax\(^{2}\) + bx + c = 0 must be satisfied by x = p + √q.

Therefore,

a(p + √q)\(^{2}\) + b(p + √q) + c = 0

⇒ a(p\(^{2}\) + q + 2p√q) + bp + b√q + c = 0

⇒ ap\(^{2}\) - aq + 2ap√q + bp + b√q + c = 0

⇒ ap\(^{2}\) - aq + bp + c + (2ap + b)√q = 0

⇒ ap\(^{2}\) - aq + bp + c + (2ap + b)√q = 0 + 0 √q

Therefore,

ap\(^{2}\) - aq + bp + c = 0 and 2ap + b = 0

Now substitute x by p - √q in ax\(^{2}\) + bx + c we get,

a(p - √q)\(^{2}\) + b(p - √q) + c

= a(p\(^{2}\) + q - 2p√q) + bp - p√q + c

= ap\(^{2}\) + aq - 2ap√q + bp - b√q + c

= ap\(^{2}\) + aq + bp + c - (2ap + b)√q

= 0 - √q 0 [Since, ap\(^{2}\) - aq + bp + c = 0 and 2ap + b = 0]

= 0

Now we clearly see that the equation ax\(^{2}\) + bx + c = 0 is satisfied by x = (p - √q) when (p + √q) is a surd root of the equation ax\(^{2}\) + bx + c = 0. Therefore, (p - √q) is the other surd root of the equation ax\(^{2}\) + bx + c = 0.

Similarly, if (p - √q) is a surd root of equation ax\(^{2}\) + bx + c = 0 then we can easily proved that its other surd root is (p + √q).

Thus, (p + √q) and (p - √q) are conjugate surd roots. Therefore, in a quadratic equation surd or irrational roots occur in conjugate pairs.


Solved example to find the irrational roots occur in conjugate pairs of a quadratic equation:

Find the quadratic equation with rational coefficients which has 2 + √3 as a root.

Solution:

According to the problem, coefficients of the required quadratic equation are rational and its one root is 2 + √3. Hence, the other root of the required equation is 2 - √3 (Since, the surd roots always occur in pairs, so other root is 2 - √3.

Now, the sum of the roots of the required equation = 2 + √3 + 2 - √3 = 4

And, product of the roots = (2 + √3)( 2 - √3) = 2\(^{2}\) - (√3)\(^{2}\) = 4 - 3 = 1

Hence, the equation is

x\(^{2}\) - (Sum of the roots)x + product of the roots = 0

i.e., x\(^{2}\) - 4x + 1 = 0

Therefore, the required equation is x\(^{2}\) - 4x + 1 = 0.





11 and 12 Grade Math 

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