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प्रश्न
Solve the following quadratic equation for x:
`x^2+(a/(a+b)+(a+b)/a)x+1=0`
उत्तर
Given:
`x^2+(a/(a+b)+(a+b)/a)x+1=0`
Let `a/(a+b)`be t
Thus, the equation becomes
`x^2+(t+1/t)x+1=0`
⇒x2+(t+`1/t`)x+1=0
⇒x2+(`(t^2+1)/1`)x+1=0
⇒tx2+(t2+1)x+t=0
⇒tx2+t2x+x+t=0
⇒(tx2+t2x)+(x+t)=0
⇒tx(x+t)+1(x+t)=0
⇒(tx+1)(x+t)=0
⇒tx+1=0, x+t=0
`=>x=(-1)/t,x= -t`
`=>x = (-1)/(a/(a+b)), x = -a/(a+b)`
`=>x= -(a+b)/a, x= -a/(a+b)`
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