Question

3-root5 upon 3+2root 5 = -19/11+a root5​

Answer 1

Answer:

–9/11

Step-by-step explanation:

As per the provided information in the given question, we have to solve for a. We have been given that,

$\twoheadrightarrow \sf{\quad {\dfrac{3-\sqrt{5}}{3+2\sqrt{5}} = -\dfrac{19}{11} + a\sqrt{5}}}$

In order to solve this question, firstly, we need to rationalise the denominator of the fraction in LHS. In order to do that so, we'll multiply the rationalising factor of the denominator with both the numerator and the denominator.

Here, the denominator is in the form of (a + b). And, the rationalising factor of (a + b) is (a – b). Thus, rationalising factor of (3 + 2√5) is (3 – 2√5).

So, we'll multiply (3 – 2√5) with both the numerator and the denominator.

$\twoheadrightarrow \sf{\quad {\dfrac{3-\sqrt{5}}{3+2\sqrt{5}} \times \dfrac{3-2\sqrt{5}}{3-2\sqrt{5}} }}$

As we know the fraction rules that a/b × c/d is equivalent to ac/bd, thus applying this rule, we can rearrange the terms.

$\twoheadrightarrow \sf{\quad {\dfrac{(3-\sqrt{5})(3-2\sqrt{5})}{(3+2\sqrt{5})(3-2\sqrt{5})} }}$

In the numerator, we have to use distributive property i.e, we have to multiply each term of the first binomial with each term of the another binomial, something like (a + b)(c + d) = ac + ad + bc + bd. Whereas, in the denominator, we'll apply the identity that is, (a + b)(a – b) = a² – b².

$\twoheadrightarrow \sf{\quad {\dfrac{3(3-2\sqrt{5})-\sqrt{5}(3-2\sqrt{5})}{(3)^2-(2\sqrt{5})^2} }}$

Performing multiplication in the numerator of the fraction. And, in the denominator, writing the squares of the terms.

$\twoheadrightarrow \sf{\quad {\dfrac{9-6\sqrt{5}-3\sqrt{5}+10}{9-20} }}$

Now, add the like terms in the numerator. And, take √5 common in –6√5 and –3√5 in the numerator. Also, performing subtraction in the denominator.

$\twoheadrightarrow \sf{\quad {\dfrac{19+ \sqrt{5}(-6-3)}{-11} }}$

Performing addition in the brackets in the numerator.

$\twoheadrightarrow \sf{\quad {\dfrac{19+ \sqrt{5}(-9)}{-11} }}$

As we know that when positive value is multiplied with the negative value, the result comes in the negative value. So, √5 times –9 will be –9√5.

$\twoheadrightarrow \sf{\quad {-\dfrac{19-9 \sqrt{5}}{11} }}$

Using another fraction rule that is, (a + b)/c = a/c + b/c, the above fraction can ne written as :

$\twoheadrightarrow \quad\boxed{\sf {-\dfrac{19}{11}+ \dfrac{-9}{11}\sqrt{5} }}$

Compare both LHS and RHS,

$\twoheadrightarrow \quad{\sf {-\dfrac{19}{11}+ \boxed{\sf\dfrac{-9}{11}}\sqrt{5}= -\dfrac{19}{11} + \boxed{\sf a}\sqrt{5} }}$

Therefore, the value of a is –9 / 11.