Question

without solving the following quadratic equation find the value of p for which the given equation has real and equal roots x^2 + (p-3)x +p =0​

Answer 1

Answer:x2 + (p - 3) x + p = 0

Here, A = 1, B = (p - 3), C = p

Since, the roots are real and equal, D = 0

B2 - 4ac = 0

(p - 3)2 - 4(1) (p) = 0

p2 + 9 - 6p - 4p = 0

p2 - 10p + 9 = 0

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

p = 1 or p = 9

Step-by-step explanation:

Answer 2

Answer:

here's ur answer dude

Step-by-step explanation:

From equation (i),

px

2

−4x+3=0

a=p,b=−4 and c=3

If the roots of this equation are equal and real, then D=0.

Therefore,

D=b

2

−4ac=0

(−4)

2

−4×p×3=0

16−12p=0

12p=16

p=

3

4

So, the value of p is

3

4

.

From equation (ii),

x

2

+(p−3)x+p=0

a=1,b=(p−3) and c=p

If the roots of this equation are equal and real, then D=0.

Therefore,

D=b

2

−4ac=0

(p−3)

2

−4×1×p=0

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−1)(p−9)

p=1,9

So, the values of p is 1,9.

hope it helps