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प्रश्न
Find the value of p for which the quadratic equation `(2p+1)x^2-(7p+2)x+(7p-3)=0` has real and equal roots.
उत्तर
The given equation is `(2p+1)x^2-(7p+2)x+(7p-3)=0`
This is of the form `ax^2+bx+c=0` where a=`2p+1, b=-(7p-2) and c=7 p-3`
∴ `D=b^2-4ac`
=`-[-7p+2]^2-4xx(2p+1)xx(7p-3)`
=`(49p^2+28p+4)-4 (14p^2+p-3)`
=`49p^2+28p+4-56p^2-4p+12`
=`-7p^2+24p+16`
The given equation will have real and equal roots if D = 0.
∴ `-7p^2+24p+16=0`
⇒ `7p^2-24p-16=0`
⇒ `7p^2-28p+4p-16=0`
⇒`7p(p-4)+4(p-4)=0`
⇒`(p-4) (7p+4)=0`
⇒` p-4=0 or 7p+4=0`
⇒ `p=4 or p=-4/7`
Hence, 4 and `-4/7` are the required values of p.
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