Can you help me simplify these double radicals?

1 answer:

Montano1993 [528]3 years ago

8 0

You know that
.. (a +b)^2 = a^2 +b^2 +2ab
If these are square roots, then you have
.. (√a +√b)^2 = a +b +2√(ab)
.. (√a -√b)^2 = a +b -2√(ab)

In each case, you can put the inner radical in the form 2√something, then look for two factors of "something" that add to the other constant.
The result is then √factor1 ±√factor2, with the sign matching that under the original radical.

1. 1 +√5
2. 2 +√2
3. √3 +√7
4. √5 -√2 . . . . . . . . Technically, √(x^2) = |x|. By expressing the result as a positive number (smaller subtracted from larger), we don't have to worry about absolute value.

5. See 3 and 4. √7 -√3
6. √20 = 2√5. See 1 and 4.
.. √5 -1
7. 2 -√3
8. 4√3 = 2√12 = 2√(6*2)
.. √6 +√2
9. 6√3 = 2√(9*3)
.. 3 -√3
10. 8√3 = 2√(16*3)
.. 4 +√3

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Write a quadratic equation with the given roots. Write the equation in the form of ax^2+bx+c=0 where a b and c are integers

Bas_tet [7]

You haven't provided the required roots, but I can tell you how to do this kind of exercises in general.

If the $x^2$ coefficient is 1, i.e. the equation is written like $x^2+bx+c=0$ , then you can say the following about the coefficients b and c:

$b$ is the opposite of the sum of the roots
$c$ is the multiplication of the roots.

So, for example, if we want an equation whose roots are 4 and -2, we have:

$4+(-2) = 4-2 = 2 \implies b = -2$
$4 \cdot (-2) = -8 \implies c = -8$

So, the equation is $x^2-2x-8=0$

If your roots are rational, you can work like this: suppose you want an equation with roots 3/4 and 1/2. You have:

$\dfrac{3}{4}+\dfrac{1}{2} = \dfrac{3}{4}+\dfrac{2}{4} = \dfrac{5}{4} \implies b = -\dfrac{5}{4}$
$\dfrac{3}{4} \cdot \dfrac{1}{2} = \dfrac{3}{8} \implies c = \dfrac{3}{8}$