How do you solve #sqrt( x -10) = 5x#?

Answer 1

There are no Real solutions,
if Complex values are allowed #x=(1+-isqrt(999))/50#

Given #color(white)("XXX")sqrt(x-10)=5x#

Working in the domain of real numbers we must ensure that

#x>=10# and #x>=0# hence finally #x>=10#.
Squaring both sides: #color(white)("XXX")x-10=25x^2#
Rewriting in standard quadratic form: #color(white)("XXX")25x^2-x+10=0#
Applying the quadratic formula #color(white)("XXX")x=(1+-sqrt(-999))/50#

Because Real numbers do not permit square roots of negative values there are no Real solutions.

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Answer 2

We have complex roots only which are #x=1/50+-isqrt(999)/50#

Squaring the two sides in #sqrt(x-10)=5x#, we get
#x-10=25x^2#
or #25x^2-x+10=0#

according to quadratic formula, the solution of the

equation #ax^2+bx+c=0# is given by
#x=(-b+-sqrt(b^2-4ac))/(2a)#.
Hence, solution is #x=(-(-1)+-sqrt((-1)^2-4*25*10))/(2*25)#
or #x=(1+-sqrt(1-1000))/(50)=(1+-sqrt(-999))/50#

As discriminant is negative, we only have complex roots,

which are #x=(1+-isqrt(999))/50#
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Answer 3

To solve the equation sqrt(x - 10) = 5x, we can follow these steps:

  1. Square both sides of the equation to eliminate the square root: (sqrt(x - 10))^2 = (5x)^2.
  2. Simplify the equation: x - 10 = 25x^2.
  3. Rearrange the equation to bring all terms to one side: 25x^2 - x + 10 = 0.
  4. This is now a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula.
  5. If we factor the quadratic equation, we need to find two numbers whose product is 10*25 = 250 and whose sum is -1. After some trial and error, we find that -10 and 25 satisfy these conditions.
  6. Rewrite the equation factored: (5x - 10)(5x + 25) = 0.
  7. Set each factor equal to zero and solve for x: 5x - 10 = 0 or 5x + 25 = 0.
  8. Solve each equation separately:
    • For 5x - 10 = 0, add 10 to both sides and divide by 5: 5x = 10, x = 2.
    • For 5x + 25 = 0, subtract 25 from both sides and divide by 5: 5x = -25, x = -5.
  9. Therefore, the solutions to the equation sqrt(x - 10) = 5x are x = 2 and x = -5.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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