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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