# 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 $36x^4 – 36x^2 + 9$.

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

So, the square root of the expression will be coming as $x^2$ because $\sqrt{x^4}= x^2$. Here the coefficient attached with $x^4$ is 36. So, let us start the division with $6x^2$ first as $\sqrt{36x^4}= 6x^2$.

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

Now we have to multiply $12x^2 – 3$ by 3 to get $-36x^2 + 9$.

The remainder turned to zero and the given expression fully divisible by $6x^2 – 3$.

So, the square root of $36x^4 – 36x^2 + 9$ will be $6x^2 – 3$.

### 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:

Find the square root of $(b + \frac{1}{b})^2 + 4(b + \frac{1}{b}) +4$

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.

How To Solve Algebraic Expressions Step By Step

Linear equations in Algebra

## 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,

$(\frac{-b}{a(1+x)})^2 = \frac{c}{ax}$

=> $(\frac{b^2}{a^2(1+x)}) = \frac{c}{ax}$

=> 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, -2.

Question 3:

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

Solution:

We know, |x| = ±x

[|x| = x for x>0 and |x| =-x for x<0 and |x| = 0 at x = 0]

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  2)/( 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, 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.

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