Question

Find the values for p for which the equation has real and equal roots x³+(p-3)x+p

Answer 1

${x}^{3} + (p - 3)x + p$

For real and equal roots,D=0.

We know that D=

${b}^{2} - 4ac = 0$

b=p-3

a=1

c=p

So

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

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

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

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

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

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

p=9,1