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प्रश्न
`15x^2-28=x`
उत्तर
Given:
`15x^2-28=x`
⇒`15x^2-x-28=0`
On comparing it with `ax^2+bx+c=0` we get;
`a=25,b=-1 and c=-28`
Discriminant D is given by:
`D=(b^2-4ac)`
=`(-1)^2-4xx15xx(-28)`
=`1-(-1680)`
=`1+1680`
=`1680`
=`1681>0`
Hence, the roots of the equation are real.
Roots α and β are given by:
`α=(-b+sqrt(D))/(2a)=(-(-1)+sqrt(1681))/(2xx25)=(1+41)/30=42/30=7/5`
`β=(-b-sqrt(D))/(2a)=(-(-1)-sqrt(1681))/(2xx25)=(1-41)/30=-40/30=(-4)/3`
Thus, the roots of the equation are `7/5` and `-4/3`
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