How to Find Square Root of Algebraic Expressions

Algebraic Expression Introduction: An algebraic expression is an expression which should include at least one variable connected by at least a single arithmetic operator. These operators can be addition(+), subtraction(-), multiplication(*) or division(/).

For example- 2 × (3 + 5) = 16 is an arithmetic expression but not an algebraic one as it doesn’t include any variable.

Square root or the square of algebraic expressions are also algebraic expressions only, and the same case happens with arithmetic expressions also.

Methods to Find Square Root of  Algebraic Expressions

The square root of algebraic expressions can be solved by two methods:

The Factoring Method

The Division Method

Finding out Square Root by Division Method

By the division method, we usually mean the long division method.

In this method, we have to see the highest power of the variable in the polynomial given. Let us see by an example:

The expression is given as 36x⁴ – 36x² + 9 .

Now, the highest power of the variable, x, is 4.

So, the square root of the expression will be coming as x² because;

As 6x² is the only term present here or it is the monomial, by multiplying by 2, we get 12x². Thus, the first term now to be divided will be -36x^2, and on dividing -36x² by 12x^2, we get -3.

Now we have to multiply 12x² – 3 by 3 to get -36x² + 9.

The remainder turned to zero, and the given expression is divisible by 6x² – 3.

So, the square root of 36x⁴ – 36x² + 9 will be 6x² – 3.

Finding out Square Root by Factoring Method

By factoring method, we usually use the factor theorem. Factor theorem states that if an algebraic expression f(x) is divisible by x = p and the remainder comes out to be zero, then, f(p) = 0 or (x – p) will be a factor of f(x).

The factor theorem is just the opposite of the remainder theorem, and we can say that if f(x) will be divisible by (x – p), then the remainder will be f(p).

Let us take an example:

Solution: Consider b + (1/b) as a.

Substituting the value of “a” in the given equation, we get a² + 4a + 4.

The given equation is in the form: (a + 2)² and one of its factor is (a + 2)

So, the square root of the given expression be b + (1/b) + 2.

Also read

How To Solve Algebraic Expressions Step By Step

Linear equations in Algebra

Quadratic equation

Solved Examples

Question 1:

If one root of ax² + bx + c is x times the other, then write the roots in terms of a,b and c.

Solution:

Let us take the roots as α and β

So, let us take β as x times α,

⇒ β = αx

We know,

α + β = -b/a and αβ = c/a

Putting the value of αx in place of β we get,

α + αx = -b/a and α × αx = c/a

⇒ α(1+x) = -b/a and α² x = c/a

⇒ α =-b/(a(1+x)) and α² = c/ax

Squaring the first term and equating with the second one we get,

⇒ b² × a × x = a² × (1+x) × c

⇒ xb² = ac(1 + x)

So, this is the condition for x in terms of a, b and c.

Question 2:

Find the roots of the equation x³ – 3x + 2.

Solution:

Let f(x) = x³ – 3x + 2

By arranging the given equation, we get,

x³ – 3x + 2 = x² (x – 1) + x(x – 1) – 2(x – 1)

= (x² + x – 2)(x – 1)

= (x + 2)(x – 1)(x – 1)

= (x + 2)(x – 1)²

To find roots f(x), put f(x) = 0

⇒ (x + 2)(x – 1)² = 0

The three roots of the given equation will be 1, 1, and -2.

Question 3:

Find the real roots of the equation x² – 6|x| + 8 = 0

Solution:

We know, |x| = ±x

By putting the negative value of x, we get:

x² – 6(-x) + 8 = 0

⇒ x² + 6x + 8 = 0

⇒ x² + 2x + 4x + 8 = 0

⇒ x(x + 2) + 4(x + 2) = 0

⇒ (x + 4)(x + 2) = 0

So, the roots of this equation will be -2, -4.

Again by taking the positive value of x, we get:

x² – 6x + 8 = 0

⇒ x² – 2x – 4x + 8 = 0

⇒ x(x – 2) – 4(x – 2) = 0

⇒ (x – 4)(x – 2) = 0

The roots of this equation will be 2, 4.

Therefore, the total roots of the equation x² – 6|x| + 8 = 0 will be ±2, ±4.

Question 4: Find the value of p if the equation 3x² – 2x + p = 0 and 6x² – 17x + 12 = 0 have a common root.

Solution:

Let a be the common root of the equations.

Then a will satisfy both the equations.

Thus,

3a² -2a + p = 0 …(i)

6a² -17 a + 12 = 0…(ii)

solving (ii) by quadratic formula, we get a = 3/2 and a = 4/3

Substitute the value of a in (i) to get p.

when a = 3/2, we get p = -15/4

when a = 4/3, we get p = -8/3

Question 5:  If a, b, c belong to R and equations ax² + bx + c = 0 and x² + 2x + 9 = 0 have a common root, show that a:b:c = 1:2:9

Solution:

The roots of the equation x² + 2x + 9 = 0 are imaginary as the discriminant D= 4 – 36 = -32 , which is negative.

As the imaginary roots of a quadratic equation are conjugate to each other, so both the roots of the equations will be common.

So, a/1 = b/2 = c/9

or, a:b:c = 1:2:9

Question 6: If sum of the roots of the equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, show that bc² , ca² , ab² are in AP.

Solution:

Let α and β be the roots of the equation ax² + bx + c = 0 .

Then, α + β= -b/a        ………..(1)

α β= c/a                     ………….(2)

Now, α + β = 1/α ² + 1/β²

or, α + β = (α²  +β²)/( α β)²

or, α + β =( (α +β)² – 2αβ ) / ( α β)²     ……..(3)

Now put the values of (α+ β)  and (α β) from (1) and (2)  in equation (3).

Thus, we get

-b/a  = (b² -2ac)/ c²

or,   -bc² = b²a -2ca²

or,    bc² +  ab² = 2ca²

Hence bc², ca², and ab² are in A.P.

Question 7: If α and β are the roots of the equation x² + px + q = 0 and x²ⁿ + pⁿ xⁿ + qⁿ = 0, where n is an even integer, prove that α/β, β/α are the roots of the equation xⁿ +1 + (x+1)ⁿ = 0.

Solution:

α  and β are the roots of the equation x² + px + q = 0

So, α + β =-p………(1)

α β =q  ……………..(2)

Since α  and β are the roots of the equation  x²ⁿ + pⁿ xⁿ + qⁿ =0

α²ⁿ + pⁿ αⁿ + qⁿ = 0

or,( αⁿ )² + pⁿ αⁿ + qⁿ = 0……….(3)

and,

( βⁿ )² + pⁿ βⁿ + qⁿ = 0…………..(4)

From (3) and (4) we see that  α/ β  and   β/ α  are the roots of y² + pⁿ y + qⁿ = 0

So, αⁿ + βⁿ = -pⁿ………(5)

α ⁿβⁿ = qⁿ   ……………..(6)

From (1), we have

α + β = -p

or, (α + β)ⁿ  = (-p)ⁿ = pⁿ  (n is even)

or, (α + β)ⁿ  = -(-pⁿ ) = -(αⁿ + βⁿ )  (from 5)

or, αⁿ + βⁿ + (α + β)ⁿ  = 0     ……..(7)

Dividing (7) by αⁿ , we get

(α/ β)ⁿ + 1 + ((α/ β) +1 )ⁿ = 0 ………(8)

Dividing (7) by  βⁿ , we get

( β/ α)ⁿ + 1 + (( β/α) +1 )ⁿ = 0 ………(9)

From (8) and (9), we see that α/ β and β/ α  are the roots of xⁿ + 1 + (x + 1)ⁿ = 0.