Here, we consider equations of the type which contain the unknown under
the radical sign and the value under the radical sign is known as radicand.
`text(Note)`
1. If roots are all even `(i.e. sqrtx , root4x , root6x, .........` etc) of an Hquation are arithmetic. In
other words, if the radicand is negative (i.e. x < 0), then the root is
imaginary, if the radicand is zero, then the root is also zero and :if the
radicand is positive, then the value of the root is also positive.
2. Ifroots are all odd `(i.e. root3x , root5x , root7x, .........` etc) of an equation, then it is defined for
all real values of the radicand. If the radicand is negative, then the root is
negative, if the radicand is zero, then the root is zero and if the radicand is
positive, then the root is positive.
`text(Some Standard Formulae to Solve Irrational Equations)`
If f and g be functions of `x, k in N.` Then,
`1. root(2k)froot(2k)g = root(2k)(fg) , f >= 0 , g >= 0 `
`2. root(2k)f//root(2k)g = root(2k)(f//g) , f >= 0 , g >= 0 `
`3. |f|root(2k)g = root(2k)(f^(2k)g) , g >= 0`
`4.root(2k)(f//g) = root(2k)|f| // root(2k)|g| . fg >= g ne 0`
`5.root(2k)(fg) = root(2k)|f|root(2k)g, fg >= 0`
`text(Some Standard Forms to Solve Irrational Equations)`
Form 1 An equation of the form
`f^(2n) (x) = g^(2n) (x), n in N` is equivalent to `f(x) = g(x).`
Then, find the roots of this equation. If root of this equation satisfies the
original equation, then its root of the original equation, otherwise, we say that
this root is its extraneous root.
`text(Squaring an Equation May Give Extraneous Roots)`
Squaring should be avoided as for as possible. If squaring is necessary, then
the roots found after squaring must be checked whether they satisfy the original
equation or not. If some values of x which do not satisfy the original equation.
These values of x are called extraneous roots and are rejected.
`text(illustration)` Solve the equation `sqrtx = x- 2 ` .
Solution. We have `sqrtx = x- 2 ` .
On squaring both sides, we obtain
`x= (x-2)^3`
`=> x^2 - 5x + 4 = 0 => (x-1) (x-4) = 0`
`therefore x_1 = 1` and `x_2 =4`
Hence, `x_1 = 4` satisfies the original equation, but `x_2 = 1` does not satisfy the
original equation.
`therefore x_2 = 1` is the extraneous root.
Here, we consider equations of the type which contain the unknown under
the radical sign and the value under the radical sign is known as radicand.
`text(Note)`
1. If roots are all even `(i.e. sqrtx , root4x , root6x, .........` etc) of an Hquation are arithmetic. In
other words, if the radicand is negative (i.e. x < 0), then the root is
imaginary, if the radicand is zero, then the root is also zero and :if the
radicand is positive, then the value of the root is also positive.
2. Ifroots are all odd `(i.e. root3x , root5x , root7x, .........` etc) of an equation, then it is defined for
all real values of the radicand. If the radicand is negative, then the root is
negative, if the radicand is zero, then the root is zero and if the radicand is
positive, then the root is positive.
`text(Some Standard Formulae to Solve Irrational Equations)`
If f and g be functions of `x, k in N.` Then,
`1. root(2k)froot(2k)g = root(2k)(fg) , f >= 0 , g >= 0 `
`2. root(2k)f//root(2k)g = root(2k)(f//g) , f >= 0 , g >= 0 `
`3. |f|root(2k)g = root(2k)(f^(2k)g) , g >= 0`
`4.root(2k)(f//g) = root(2k)|f| // root(2k)|g| . fg >= g ne 0`
`5.root(2k)(fg) = root(2k)|f|root(2k)g, fg >= 0`
`text(Some Standard Forms to Solve Irrational Equations)`
Form 1 An equation of the form
`f^(2n) (x) = g^(2n) (x), n in N` is equivalent to `f(x) = g(x).`
Then, find the roots of this equation. If root of this equation satisfies the
original equation, then its root of the original equation, otherwise, we say that
this root is its extraneous root.
`text(Squaring an Equation May Give Extraneous Roots)`
Squaring should be avoided as for as possible. If squaring is necessary, then
the roots found after squaring must be checked whether they satisfy the original
equation or not. If some values of x which do not satisfy the original equation.
These values of x are called extraneous roots and are rejected.
`text(illustration)` Solve the equation `sqrtx = x- 2 ` .
Solution. We have `sqrtx = x- 2 ` .
On squaring both sides, we obtain
`x= (x-2)^3`
`=> x^2 - 5x + 4 = 0 => (x-1) (x-4) = 0`
`therefore x_1 = 1` and `x_2 =4`
Hence, `x_1 = 4` satisfies the original equation, but `x_2 = 1` does not satisfy the
original equation.
`therefore x_2 = 1` is the extraneous root.