6x^2+2x-3/2=0 (the 6x^2 and 3/2 are square roots)

Question

Trying to work this problem but not sure where to start.

Answer 1

Hi CiJi,

Let's begin by writing out our equation as you indicated

(the 6x^2 and -3/2 terms being under the square root sign):  sqrt(6x^2) + 2x - sqrt(3/2) = 0

Now let's isolate the sqrt(6x^2) term on the left side by moving the other two terms to the right side (reversing their signs on the right side) - this will cancel out these terms on the left side:  -2x cancels out +2x and plus sqrt(3/2) cancels out minus sqrt(3/2):  sqrt(6x^2) = -2x + sqrt(3/2)

Now let's get rid of the radical (square root sign) on the left side by squaring both sides of the equation remembering that on the right side - squaring a binomial expression is the same as multiplying the binomial expression by itself:  6x^2 = (-2x + sqrt(3/2))(-2x + sqrt(3/2))

Now let's multiply the two binomial expressions on the right side by multiplying each term in the 1st set of parens by each term in the 2nd set of parens:  

6x^2 = (-2x)(-2x) + (-2x)(sqrt(3/2) + (sqrt(3/2))(-2x) + (sqrt(3/2))(sqrt(3/2))

Now let's write out the multiplication results on the right side:

6x^2 = 4x^2 - 2sqrt(3/2)x - 2sqrt(3/2)x + 3/2

Now let's simplify the right side by combining like terms:  6x^2 = 4x^2 - 4sqrt(3/2)x + 3/2

Now let's set the right side to zero and move the terms on the right side to the left side by reversing the sign of each term to cancel out these terms on the right side: 

6x^2 - 4x^2 + 4sqrt(3/2)x - 3/2 = 0

Now let's simplify the left side by combining like terms:  2x^2 + 4sqrt(3/2)x - 3/2 = 0 

We now have the standard form of a quadratic equation (ax^2 + bx + c = 0): 

2x^2 + 4sqrt(3/2)x - 3/2 = 0 

Our coefficients (a, b, c) are:  a = 2 and b = 4sqrt(3/2) and c = -3/2

We can now use the quadratic formula to solve for x:  x = (-b +/- sqrt(b^2 - 4ac)) / 2a

Plugging in our coefficient values for a and b and c:

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

Squaring b coefficient (4sqrt(3/2)^2): 

x = (-4sqrt(3/2) +/- sqrt(16(3/2) - 4(2)(-3/2))) / 2(2)

Performing all multiplication operations:

x  = (-4sqrt(3/2) +/- sqrt(24 + 12)) / 4

Performing addition operation under square root sign:

x = (-4sqrt(3/2) +/- sqrt(36)) / 4

Extracting perfect square (taking square root of 36):

x = (-4sqrt(3/2) +/- 6) / 4

Final simplification (dividing each term in numerator and denominator by 4 to get rid of fraction):

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

Final Exact Answers for x in radical form: 

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

x = -sqrt(3/2) - 3/2

Final Approximate Answers for x in decimal form rounded to nearest hundredth:

x = -sqrt(3/2) + 3/2 = -1.22 + 1.50 = 0.28

x = -sqrt(3/2) - 3/2 = -1.22 - 1.50 = -2.72

Whew !! 

This was definitely a challenging problem, but I trust the above stepwise explanation will be helpful to you.

Thanks for submitting your question.

Regards, Jordan.

Answer 2

Alright, I am going to assume that the equation should look like this:

√(6x²) + 2x - √(3/2)=0

If that is the case, then you have to solve a radical equation. You can do that by isolating √(6x²) from the other terms like this: 

√(6x²) = √(3/2) - 2

Then you proceed by raising both sides (note: sides, not individual terms) to the power of two: 

(√(6x²))² = (√(3/2) - 2x)²

This allows to get rid of the square root from √(6x²) since the square root is equal to the power of 1/2

6x² = (√(3/2) - 2x)²

6x² = 3/2 - 2√(3/2)2x +4x²

6x² - 4x² + 4√(3/2)x -3/2 = 0

2x² + 4√(3/2)x - 3/2 = 0

Now that is a quadratic equation of the form ax² + bx + c = 0. Therefore we apply the quadratic formula: x= [-b ± √(b²-4ac)] / 2a

which gives us

x = {-4√(3/2) ± √[4√(3/2)² - 4(2)(3/2)] } / 4

Remember that a square root is the same thing as the power 1/2 therefore,

4√(3/2)² = 4²(3/2)^1/2*2 = 16(3/2)¹ = 16(3/2) = 24

So,

x= [-4√(3/2) ± √(24 - 12)] / 4 

x= [-4√(3/2) ± √12]/4

x= -√(3/2) ± √12/4 

This gives you two solutions for x

x₁= -√(3/2) + √12/4 ≈ -0.358719467607589 

x₂= -√(3/2) - √12/4 ≈ -2.090770275175589

I hope this helps! Let me know if what I started with wasn't the correct equation, I was a bit confused by the instructions