Solving Equations with Radicals and Exponents

Solving equations with radicals and exponents:

Exponents:      

The term exponent in math is used to find the exponential value of the particular value it may be integer  or fraction . We can easy to get  the power of a  number using online calculator ,Consider  the unknown number , Here x is base and y is the power  of x . Let us consider the  number 23 Here 2 is the base(x)  and 3 is the power of 2(y). We calculate  23= 2 x 2 x 2 =8

Radicals:

Square root of the number is said to be a radical number .A radical equation is an equation in which a variable is under a radical term.Let us discuss about the solving an equations with radical and  exponents,

Example Problems to Solving Equations with Radicals and Exponents:

Example 1: Solve `sqrt(3x^2 +8x)` -3 =0

Solution:

Given `sqrt(3x^2+8x)` -3=0

isolate the radical equation,

`sqrt(3x^2 +8x)=3`

Take square on both sides we get,

`(sqrt3x^2+8x)^2` `=3^2`

3x^2+8x=9

3x^2+8x-9 =0

now you can solve the above equation using factoring method,

Using quadratic formula ,

x= `(-b+-sqrt(b^2-4ac))/(2a)`

Here b=8 ,a=3 and c=-9

Therefore ,

x= `(-8+-sqrt(8^2-(4)(3)(-9)))/(2(3))`

`=(-8+-sqrt(64+108))/6`

`=(-8+-sqrt(172))/6`

`=(-8+-sqrt(2*2*43))/6`

`=(-8+-2sqrt(43))/6`

`=2([-4+-sqrt(43)])/6`

`=(-4+-sqrt43)/3`

Therefore,

The factors are , `(-4+sqrt43)/3` ,`(-4-sqrt43)/3`

Example 2 : Solve `sqrt( x^2 +5x)-3=0`

Solution:

Isolate the given radical and exponent equation, we get,

`sqrt(x^2+5x)=3`

Take square root on both sides we get,

x^2+5x=32

x^2+5x=9

x^2+5x-9=0

Above equation in the form of ax^2+bx+c.

Therefore  you can find the factors using quadratic formula,

`x= (-b+-sqrt(b^2-4ac))/(2a)`

Here , b=5,a=1 and c=-9

substitute these values into formula,

`x= ((-5+-sqrt(5^2-4(1)(-9)))/(2(1)))`

`=(-5+-sqrt(25+36))/2`

`=(-5+-sqrt(61))/2`

`=(-5+-sqrt(61))/2`

Therefore ,

The factors are ,

`x= (-5+sqrt61)/2`    ,   `(-5-sqrt61)/2`

More about the Solving of Equations with Radicals and Exponents:

Example3: Solve `sqrt(2x^2+4)=4`

Solution:

Take square on both sides we get,

`(sqrt(2x^2+4))^2` =42

2x^2+4 =16

subtract both sides by 4,we get

2x^2+4-4=16-4

2x^2=12

Divide both sides by 2 ,

x^2 =`12/2`

x^2=6

x=`sqrt6`

Example 4: Solve the equations with radicals  and exponents `sqrt(4x^2+8)=11`

Solution:

Take square on both sides , we get,

4x^2+8 =121

Subtract both sides by 8 we get,

4x^2+8-8 =121-8

4x^2=113

Divide both side by 4,

x^2 =113/4

x=`sqrt(113/4)`

x=`(sqrt113)/2`

Therefore the value of `x= sqrt113/2`

Example 5: Solve the equation with radicals and exponents `sqrt(6x^2+7)` `=8`

Solution:

Take square on both sides we get,

6x^2 +7 = 64

Subtract both side by7,

6x^2+7-7=64-7

6x^2=57

Divide both side by 6,

x^2 =`57/6`

x= `sqrt(57/6)`

Therefore the value of `x = sqrt(57/6)`