Question

plzz tell the answer for class 10​

Answer 1

Question :

Find the value of p for which the equation

(p+1)x²-6(p+1)x+3(p+9)=0, p≠-1 has equal real roots.

Theory :

For a Quadratic equation of the form

ax²+bx+c= 0 , the expression b²-4ac is called the discriminant.

Nature of roots

The roots of a quadratic equation can be of three types.

If D>0, the equation has two distinct real roots.

If D=0, the equation has two equal real roots.

If D<0, the equation has no real roots.

Solution :

Given Quadratic equation :

(p+1)x²-6(p+1)x+3(p+9)=0

On comparing with standard form of quadratic equation i.e ax² + bx + c =0 ,

Here ,

• a = p+1
• b= -6(p+1)
• and c= 3(p+9)

We know that, When roots are equal

Then ,Discriminant = 0

$\implies\sf\:[6(p+1)]^2-4\times(p+1)\times3(p+9)=0$

$\implies\sf\:36(p^2+1+2p)-12\times(p+1)\times(p+9)=0$

$\implies\sf\:36(p^2+1+2p)-12(p^2+9p+p+9)=0$

$\implies\sf\:36(p^2+1+2p)-12\times(p^2+10p+9)=0$

$\implies\sf\:36p^2+36+72p-12p^2-120p-108=0$

$\implies\sf\:24p^2-48p-72=0$

$\implies\sf24(p^2-2p-3)=0$

$\implies\sf\:p^2-2p-3=0$

Now factorize

$\sf\:p^2-2p-3=0$

$\sf\:p^2-3p+p-3=0$

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

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

$\sf\:p=3\:or\:-1$

But ,given p≠-1

Thus ,p=3

Put the value of p = 3 in given equation

(p + 1)x²  - 6(p  + 1) x  + 3 (p + 9) = 0.

(3+1)x² -6(3+1)x + 3(3+9)= 0

⇒4x² -24x +36= 0

⇒4(x² -6x +9)=0

⇒x² -6x +9 = 0

⇒x² -3x -3x +9= 0

⇒x(x -3) -3(x -3)= 0

⇒(x-3)(x-3)= 0

Then , (x-3)= 0  or  (x-3)= 0

Therefore ,Roots of the equation are 3 ,3 ( equal roots )

Answer 2

we know that when the roots are equal

decriment = 0

$[6(p + 1) {]}^{2} - 4 \times (p + 1) \times 3(p + 9) = 0$

$36( {p}^{2} + 1 + 2p) - 12 \times (p + 1) \times (p + 9) = 0$

$36( {p}^{2} + 1 + 2p) - 12( {p}^{2} + 9p + p + 9) = 0$

$36( {p}^{2} + 1 + 2p) - 12 \times ( {p}^{2} + 10p + 9) = 0$

$36 {p}^{2} + 36 + 72p - 12 {p}^{2} - 120p - 108 = 0$

$24 {p}^{2} - 48p - 72 = 0$

$24( {p}^{2} - 2p - 3) = 0$

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

factorise

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

given p ≠ 3

So,P = 3

put the value of p = 3 in the equation

$(p + 1) {x}^{2} - 6(p + 1)x + 3(p + 9) = 0 \\ (3 + 1) {x}^{2} - 6(3 + 1)x + 3(3 + 9) = 0 \\ \\ = > 4 {x}^{2} - 24x + 36 = 0 \\ = > 4( {x}^{2} - 6x + 9) = 0 \\ = > {x}^{2} - 6x + 9 = 0 \\ = > {x}^{2} - 3x - 3x + 9 = 0 \\ = > x(x + 3) - 3(x - 3) = 0$

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

So (x-3) = 0 or (x-3) = 0

so the roots of equation 3,3 (equal values)