Question

find the value of p war which the quadratic equation (p+1)x² - 3(p-1) - (p-1) = 0 have two equal roots​

Answer 1

Answer:

answer for the given problem is given

Answer 2

Answer:

$\large\boxed{\star\:\:\:p=1\:,\:\frac{5}{13} \:\:\:\star}$

Step-by-step explanation:

❈ Given:-

   (p+1)x² - 3(p-1) - (p-1) = 0 have equal roots​.

❈ To Find:-

    Value of p.

❈ Formula Applied:-

•     ax²+bx+c=0, where this Quadratic Equation have Real and Equal Roots.

∴ b²-4ac=0 (Real and Equal Roots)

• (a+b)(a-b)=a²-b²

❈ Solution:-

    b²-4ac=0, where a=(p+1), b=[-3(p-1)] and c=[-(p-1)]

⇒ [-3(p-1)]²-4[-(p-1)](p+1)=0

⇒ 9(p-1)²+4(p-1)(p+1)=0

⇒ 9(p²-2p+1)+4[(p)²-(1)²]=0

⇒ 9p²-18p+9+4(p²-1)=0

⇒ 9p²-18p+9+4p²-4=0

⇒ 13p²-18p+5=0

⇒ 13p²-13p-5p+5=0

⇒ 13p(p-1)-5(p-1)=0

⇒ (13p-5)(p-1)=0

$\implies\boxed{p=1\:,\:\frac{5}{13} }$