13-a*root 10 whole root = root 8 + root 5 then a equals to ???

In your last two posts, I don't understand where the

"whole root" is supposed to go.

To find the value of "a," we need to isolate it on one side of the equation. Let's break down the steps:

Given equation: 13 - a√10√ = √8 + √5

First, we can simplify the equation by multiplying the radicals. Note that the product of two square roots gives us a square root of the product inside. Therefore:

√10√ = √(10 * 1) = √10

The equation becomes: 13 - a√10 = √8 + √5

Next, let's move the terms with square roots (radicals) to one side of the equation and the constant terms to the other side:

- a√10 = (√8 + √5) - 13

Now, let's square both sides of the equation to eliminate the square root:

(- a√10)² = (√8 + √5 - 13)²

Squaring the left side gives us: a² * 10

For the right side, we need to expand the squared term using the formula (a + b)² = a² + 2ab + b²:

(√8 + √5 - 13)² = (√8 + √5 - 13)(√8 + √5 - 13)
= (√8)² + 2(√8)(√5) + 2(√5)(-13) + (√5)² + 2(√5)(-13) + (-13)²
= 8 + 2√40 - 26√5 + 5 - 26√5 + 169
= 182 - 52√5

The equation now becomes: a² * 10 = 182 - 52√5

Finally, divide both sides by 10 to solve for "a":

a² = (182 - 52√5) / 10

Simplifying further, we have:

a² = 18.2 - 5.2√5

Taking the square root of both sides, we get:

a = ±√(18.2 - 5.2√5)

So, the value of "a" is given by: a = ±√(18.2 - 5.2√5)