If x2−4x2+4=35\dfrac{x^2 - 4}{x^2 + 4} = \dfrac{3}{5}x2+4x2−4=53, the value of x is :
4
±4\pm 4±4
14\dfrac{1}{4}41
±14\pm \dfrac{1}{4}±41
13 Likes
Given,
⇒x2−4x2+4=35⇒5(x2−4)=3(x2+4)⇒5x2−20=3x2+12⇒5x2−3x2=12+20⇒2x2=32⇒x2=16⇒x=16⇒x=±4.\Rightarrow \dfrac{x^2 - 4}{x^2 + 4} = \dfrac{3}{5} \\[1em] \Rightarrow 5(x^2 - 4) = 3(x^2 + 4) \\[1em] \Rightarrow 5x^2 - 20 = 3x^2 + 12 \\[1em] \Rightarrow 5x^2 - 3x^2 = 12 + 20 \\[1em] \Rightarrow 2x^2 = 32 \\[1em] \Rightarrow x^2 = 16 \\[1em] \Rightarrow x= \sqrt{16} \\[1em] \Rightarrow x = \pm 4.⇒x2+4x2−4=53⇒5(x2−4)=3(x2+4)⇒5x2−20=3x2+12⇒5x2−3x2=12+20⇒2x2=32⇒x2=16⇒x=16⇒x=±4.
Hence, Option 2 is the correct option.
Answered By
9 Likes
If a, b, c and d are in proportion, the value of 8a2−5b28c2−5d2\dfrac{8a^2 - 5b^2}{8c^2 - 5d^2}8c2−5d28a2−5b2 is equal to :
a2 : b2
a2 : c2
a2 : d2
c2 : d2
x+yz=y+zx=z+xy\dfrac{x + y}{z} = \dfrac{y + z}{x} = \dfrac{z + x}{y}zx+y=xy+z=yz+x is equal to :
0
1
2
-2
If xa+b−c=yb+c−a=zc+a−b\dfrac{x}{a + b - c} = \dfrac{y}{b + c - a} = \dfrac{z}{c + a - b}a+b−cx=b+c−ay=c+a−bz = 5 and a + b + c = 7; the value of x + y + z is :
35
75\dfrac{7}{5}57
57\dfrac{5}{7}75
42
If a : b = c : d, prove that :
5a + 7b : 5a - 7b = 5c + 7d : 5c - 7d
