If x = sqrt{5} + 2,then the value of frac{2x^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $x = \sqrt{5} + 2$.First,calculate $x^2 = (\sqrt{5} + 2)^2 = 5 + 4 + 4\sqrt{5} = 9 + 4\sqrt{5}$.Substitute $x$ and $x^2$ into the numerator:

Vedclass · Accepted Answer

$0.625$

Vedclass · Answer

(C) Given $x = \sqrt{5} + 2$.First,calculate $x^2 = (\sqrt{5} + 2)^2 = 5 + 4 + 4\sqrt{5} = 9 + 4\sqrt{5}$.Substitute $x$ and $x^2$ into the numerator: $2(9 + 4\sqrt{5}) - 3(\sqrt{5} + 2) - 2 = 18 + 8\sqrt{5} - 3\sqrt{5} - 6 - 2 = 10 + 5\sqrt{5}$.Substitute $x$ and $x^2$ into the denominator: $3(9 + 4\sqrt{5}) - 4(\sqrt{5} + 2) - 3 = 27 + 12\sqrt{5} - 4\sqrt{5} - 8 - 3 = 16 + 8\sqrt{5}$.Now,the expression is $\frac{10 + 5\sqrt{5}}{16 + 8\sqrt{5}}$.Factor out constants: $\frac{5(2 + \sqrt{5})}{8(2 + \sqrt{5})} = \frac{5}{8}$.Calculating the decimal value: $\frac{5}{8} = 0.625$.

If $x = \sqrt{5} + 2$,then the value of $\frac{2x^2 - 3x - 2}{3x^2 - 4x - 3}$ is equal to

Similar Questions

Find the value of $(a+b+c)^{3}-3(b+c)(c+a)(a+b)$ if $a=5, b=3$ and $c=2$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+13x+40=0$
$II.$ $y^{2}+7y+10=0$

If the equations $x^{2}+2x-3=0$ and $x^{2}+3x-k=0$ have a common root,then the non-zero value of $k$ is:

Solve the given two equations and select the correct answer from the given options.
$I. 3x^2 - 4x - 32 = 0$
$II. 2y^2 - 17y + 36 = 0$

If $\log _{10}(x^{2}-6x+45)=2$,then the values of $x$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module