Question

Find the value of p for which the quadratic equation
x²+(p - 3)x+p = 0 has real and equal root ​

Answer 1

Step-by-step explanation:

x²+(p - 3)x+p = 0 if this equation has equal root then discriminant(D) = 0

D= b²-4ac

0= (p-3)²-4×1×p

0= p²+9-6p-4p

0= p²-10p+9

0= p²-9p-p+9

0= p(p-9)-1(p-9)

0= (p-1)(p-9)

0=p-1 p-9=0

p=1 p=9

hence value of p is 1 or 9

#666

Answer 2

p=9 or P=1

Step-by-step explanation:

x²+(p - 3)x+p = 0

Hence the equation has real and equal roots so,

$(p - 3) ^{2} - 4 \times 1 \times p = 0$

(By using discriminant formula )

$p ^{2} + 9 - 6p - 4p = 0$

$p ^{2} - 10p + 9 = 0$

$p ^{2} - 9p - p + 9 = 0$

$p(p - 9) - 1(p - 9) = 0$

$(p - 9) (p - 9) = 0$

$p - 9 = 0 \: or \: p - 1 = 0$

so,

$p = 9 \: or \: p = 1$