Questions on Algebra: Radicals -- complicated equations involving roots answered by real tutors!

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12/sqrt(5x+6) = sqrt(2x+5)

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To find 'x', start by multiplying both sides by sqrt(5x+6)

(12/sqrt(5x+6))sqrt(5x+6) = sqrt(2x+5)sqrt(5x+6)

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Notice the denominator cancels on the left. On the right we have:
12 = sqrt((2x+5)(5x+6))

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Now, we square both sides:
144 = (2x+5)(5x+6)

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Using FOIL, we expand the right side:
144 = 10x^2 + 25x + 12x + 30

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Since, we can't factor, we must use the quadratic equation. Doing so, will produce two solutions:
x = {2, -5.7}
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Reference: here is the quadratic solution

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation ax^2+bx+c=0

(in our case 10x^2+37x+-114 = 0

) has the following solutons:

$x[12] = (b+-sqrt( b^2-4ac ))/2\a$

For these solutions to exist, the discriminant b^2-4ac

should not be a negative number.

First, we need to compute the discriminant b^2-4ac

.

Discriminant d=5929 is greater than zero. That means that there are two solutions: $x[12] = (-37+-sqrt( 5929 ))/2\a$ .

$x[1] = (-(37)+sqrt( 5929 ))/2\10 = 2$
$x[2] = (-(37)-sqrt( 5929 ))/2\10 = -5.7$

Quadratic expression 10x^2+37x+-114

can be factored:
10x+37x+-114 = 10(x-2)*(x--5.7)

Again, the answer is: 2, -5.7. Here's your graph:
graph( 500, 500, -10, 10, -20, 20, 10*x^2+37*x+-114 )