Question

please solve these two maths equation...​

Answer 1

SOLUTION

(i)x−2x+2+x+2x−2=4

By taking LCM, (x−2)2+(x+2)2(x+2)(x−2)=4

x2−4x+4+x2+4x+4x2−4=4

By simplifying the equation, we get

2x2+8=4x2−162x2+8−4x2+16=0−2x2+24=0x2−12=0

Let us consider,

a=1,b=0,c=−12

So, by using the formula,

x=−b±b2−4ac−−−−−−−√2a So let, b2−4ac=Dx=−b±D−−√2aD=b2−4ac=(0)2−4(1)(−12)=0+48=48

So,

x=[−(0)±48−−√]/2(1)=[±48−−√]/2=[±(√16×3)]/2=±43√/2=±23√=23√ or −23√

∴ Value of x=23√,−23√

(ii)x+1x+3=3x+22x+3

Let us cross multiply,  

we get (x+1)(2x+3)=(x+3)(3x+2)

Now by simplifying  

we get 2x2+3x+2x+3=3x2+9x+2x+6

2x2+5x+3−3x2−11x−6=0

−x2−6x−3=0

x2+6x+3=0

Let us consider a=1,b=6,c=3

So, by using the formula,

x=−b±b2−4ac−−−−−−−√2a So let, b2−4ac=Dx=−b±D−−√2aD=b2−4ac=(6)2−4(1)(3)=36−12=24

So,

x=[−(6)±24−−√]/2(1)=[−6±(4×6)]−−−−−−√/2=[−6±26√]/2=−3±6√=−3+6√ or −3−6√∴ Value of x=−3+6√,−3−6√