Question

Without solving the equation find tue value of p for which the equation (p-3)x2 -2(p-3)x+2

Answer 1

ANSWER

P = 3 & 5

EXPLANATION

1) =TO FIND value of p

2)= solution

(p-3) x^2 - 2(p-3)x +2

D =b^2 - 4ac

= > 4(p-3)^2 - 4(p-3)(2)=0

= > 4(p^2 +9 - 6p) - 8p+24 =0

= > 4p^2 + 36 -24p - 8p + 24 = 0

= > 4p^2 - 32p + 60 =0

= > p^2 - 8p + 15 =0

= > p^2 - 5p - 3p + 15 =0

= > p(p - 5) - 3(p - 5) = 0

= > (p - 3) ( p - 5) =0

= > p = 3 & p = 5

Answer 2

$\fbox\red{Question}$

Without solving the equation find tue value of p for which the equation (p-3)x2 -2(p-3)x+2??

$\fbox\red{To\:find}$

The value of p.

$\fbox\red{Solution}$

$(p - 3) {x}^{2} - 2(p - 3)x + 2 \\ \\ d = {b}^{2} - 4ac \\ \\ = > 4(p - 3)^{2} - 4(p - 3)2 = 0 \\ = > 4( {p}^{2} + 9 - 6p) - 8p + 24 = 0 \\ = > {4p}^{2} + 36 - 24p - 8p + 24 = 0 \\ = > {4p}^{2} - 32p + 60 = 0 \\ = > {p}^{2} - 5p - 3p + 15 = 0 \\ = > p(p - 5) - 3(p - 5) = 0 \\ = > (p - 3)(p - 5) = 0$

$\fbox{P\:=\:3}$

$\fbox{P\:=\:5}$