How do you solve $sqrt(10–x)+x=8$ ?

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Answer 1 · 2016-03-05T08:25:38

$x=9,6$

$sqrt(10-x)+x=8$

$rarrsqrt(10-x)=8-x$

Square both sides in order to get rid of the radical sign

$rarr(sqrt(10-x))^2=(8-x)^2$

Use the formula $(a-b)^2=a^2-2ab+b^2$

$rarr10-x=64-16x+x^2$

$rarr10=64-16x+x^2+x$

$rarr10=64-15x+x^2$

$rarr0=64-15x+x^2-10$

$rarr0=54-15x+x^2$

Rewrite

$rarrx^2-15x+54=0$

Factor the equation

$rarr(x-9)(x-6)=0$

If we solve for it we get $x=9,6$

But,

If we take the value of $9$ of $x$ into the equation,

$sqrt(10-9)+9=0$

$sqrt1+9=0$

$1+9!=0$

We should take $sqrt1$ as $-sqrt1$

Hope! I had given the right answer!

How do you solve sqrt(10–x)+x=8sqrt(10–x)+x=8?