Question

State whether the following statement is true or false.
The product of two binomial quadratic surds is always rational.
A) True
B) False

Answer 1

Hint: Before entering into the solution we must know about surds, quadratic surds and binomial quadratic surds and conjugate surds.
Surd is an irrational root of a number. A quadratic surd is an expression where the number under the square root is rational but not a square root.

Complete step-by-step answer:
Step-1
Numbers like $\sqrt 2 $,$\sqrt 3 $,$\sqrt[3]{5}$ are called surds which satisfy the conditions of being irrational, order > 1 and having positive rational numbers as base.
Surds having order 2 is called quadratic surds.
And binomial surds are those which have two terms and one of them must be surds such as $\sqrt 2 + \sqrt 3 $,$2 + 3\sqrt 5 $.
So the general form of binomial quadratic surd is, $x + y\sqrt a $.
where, x and y = non zero rational
$\sqrt a $= quadratic surd
Step-2
We are asked if the product of two binomial quadratic surds is rational or not.
Case 1- let’s take two binomial quadratic surds as $2 + 3\sqrt 5 $and$2 - 3\sqrt 5 $
Multiplying both we get,
$(2 + 3\sqrt {5)} $$(2 - 3\sqrt {5)} $
= ${2^2} - (3\sqrt {5{)^2}} $[as we know $(a + b)(a - b) = {a^2} - {b^2}$]
= $4 - 45$
= $ - 41$
-41 is a rational number.
Step-3
Case 2- Let’s take two binomial quadratic surd as $2 + 3\sqrt 5 $and $3 + 2\sqrt 6 $
Multiplying both we get,
$(2 + 3\sqrt {5)} $$(3 + 2\sqrt 6 )$
= $(2 \times 3) + (2 \times 2\sqrt {6)} + (3 \times 3\sqrt {5)} + (3\sqrt {5 \times } 2\sqrt {6)} $
= $6 + 4\sqrt {6 + } 9\sqrt 5 + 6\sqrt {30} $
This is an irrational number.
Step-4
Hence, the product of two binomial quadratic surds are not always rational.
Step-5
The statement is false. Option (B) is correct.

Note: We can also get the answer using a general form of binomial quadratic surds.
If the product of two binomial surds is a rational number then both the surds are called conjugate surds.
In case 1, both the surds were conjugate surds.