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sqrt(x^2+5) = 2x-3    solve for x

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samerrima | Student, Grade 9 | eNoter

Posted August 22, 2010 at 11:38 AM via web

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sqrt(x^2+5) = 2x-3    solve for x

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 22, 2010 at 11:42 AM (Answer #1)

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

Let us square both sides:

==> [sqrt(x^2+5)]^2 = (2x-3)^2

==> x^2 +5 = 4x^2 - 12x +9

Now group similar:

==> 3x^2-12x +4 = 0

==> x1= [12 + sqrt(144- 4*3*4)]/6= [12 + sqrt96]6 = [12+9.8]/6= 3.63

==> x2= [12- sqrt96]6 = 0.37

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thewriter | College Teacher | Valedictorian

Posted August 22, 2010 at 11:50 AM (Answer #2)

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We have to solve sqrt(x^2+5)=2x-3

First square both sides of the equation

[sqrt(x^2+5)]^2=(2x-3)^2

=>x^2+5=4x^2-12x+9

=>4x^2-x^2-12x+9-5=0

=>3x^2-12x+4=0

Now finding the roots of the above quadratic equation as [-b+sqrt(b^2-4ac)]/2a and [-b-sqrt(b^2-4ac)]/2a

here a=-12, b=3 and c=4, we get

x=-(-12)+sqrt(12^2-4*3*4)/(2*3)

=(12+sqrt96)/6=3.6329

and x=-(-12)-sqrt(12^2-4*3*4)/(2*3)

=(12-sqrt96)/6=.367

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krishna-agrawala | College Teacher | Valedictorian

Posted August 22, 2010 at 1:14 PM (Answer #3)

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Given:

Sqrt(x^2 + 5) = 2x - 3

Taking square of both the sides:

x^2 + 5 = (2x - 3)^2

==> x^2 + 5 = 4x^2 - 12x + 9

==> 3x^2 - 12x + 4 = 0

==> 3x^2 - 12x + 4 = 0

This is an equation in the  form:

ax^2 + bx + c = 0.

in which there are two possible values of x given by:

x1 = [-b + (b^2 - 4ac)^1/2]/2a

x2 = [-b - (b^2 - 4ac)^1/2]/2a

Substituting values of a, b and c in equation for x1 and x2:

x1 = (12 + (144 - 4*3*4)^1/2]/(2*3)

= [12 + (96)^1/2]/6

= (12 + 9.7979)/6

= 3.63299

x2 = (12 - (144 - 4*3*4)^1/2]/(2*3)

= [12 - (96)^1/2]/6

= [12 - 9.7979]/6

= 0.3670

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neela | High School Teacher | Valedictorian

Posted August 22, 2010 at 1:43 PM (Answer #4)

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sqrt(x^2+3) =2x-3

To solve for x .

Solution:

We can get rid of the square root  on the left by squaring both sides of the given equation and get:

x^2+3 = (2x-3)^2

x^2+3 = (2x)^2+2*(2x)(-3) +(-3)^2, as (a+b)^2 = a^2+2ab+b^2.

Simplify each term on the right.

x^2+ 5 = 4x^2-12x+9

Subtract x^2+5  from both sides.

0 = 4x^2-12x+9 - x^2-5

0 = 3x^3 -12x+4.

0 = 3x^2-12x +4.

 This is a quadratic equation of the form ax^2+bx+c  = 0, whose roots are given by:

x1 = {-b+sqrt(b^2-4ac)}/2a and x2 = {-b-sqrt(b^2-4ac)}/2a

So in 3x^2 - 12x + 2 , a = 3,  b = -12 and  c = 4.

So x1 = (- -12 +sqrt( (-12)^2- 4*3*4)}/(2*3) = (12+sqrt96)/6

x1 = (6+2sqrt6)/3.

x2 = (6-2sqrt6)/3.

 

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giorgiana1976 | College Teacher | Valedictorian

Posted August 22, 2010 at 3:03 PM (Answer #5)

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Before solving the equation, we'll set the constraint of existence of the sqrt.

x^2 + 5 >0

Now, we'll solve the equation, raising to square both sides, in order to eliminate the square root:

[sqrt(x^2+5)]^2 = (2x-3)^2

x^2+5 = 4x^2 - 12x + 9

We'll move all terms to one side and we'll combine like terms:

x^2 - 4x^2 + 12x + 5  -9 = 0

-3x^2 + 12x - 4 = 0

We'll multiply by -1:

3x^2 - 12x + 4 = 0

We'll apply the quadratic formula:

x1 = [12+sqrt(144-48)]/6

x1 = (12+sqrt96)/6

x1 = (12+4sqrt6)/6

x1 = 4(3+sqrt6)/6

x1 = 2(3+sqrt6)/3

x1 = 2 + 2sqrt6/3

x2 = 2 - 2sqrt6/3

Since both solution are positive, they are admissible, so the equation has 2 solutions.

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