Question

plsss do this sum.....
no.14
PLSSS ...DO THAT​

Answer 1

Answer:

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Answer 2

We have

$(p + 1)x {}^{2} - 6(p \: + 1)x \\ + 3(p + 9) = 0$

Compare with

$a = p + 1$

$b = - 6(p + 1)$

$c =( p + 9)$

For real and equal roots ,

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

$=36(p + 1) {}^{2} - 4(p + 1 ) \times \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3(p + 9 )= 0$

$3(p {}^{2} + 2p + 1 )- (p + 1) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (p + 9) = 0$

$3p {}^{2} \: + 6p \: + 3 - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( p {}^{2} \: + 9p \: + p \: + 9)$

$2p {}^{2} \: - 4p \: - 6 = 0$

$p {}^{2} - 2p - 3 = 0$

$p {}^{2} \: - 3p \: + p \: - 3 = 0$

$p \: (p \: - 3) + 1(p - 3 )= 0$

$(p - 3)(p + 1) = 0$

$p = - 1 \: \: and \: + 3$

Neglecting -1 we get p=3

Now the equation becomes

$4x {}^{2} - 24x + 36 = 0$

$x {}^{2} - 6x + 9 = 0$

$(x - 3)(x - 3) = 0$

x =3,3

Thus roots are 3 and 3

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