Question

If the roots of the following quadratic equation (p+1)x*2-6(p+1)x+3(p+9)=0 are equal find p and the roots of the equation

Answer 1

Answer:

here is your answer mate...

Step-by-step explanation:

Given quadratic equation is= (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, a≠ 0

Here, a = p+1 , b= -6(p+1) , c= 3(p+9)

D(discriminant)= b²-4ac

=-6 (p+1)² - 4× (p+1)× 3(p+9)

= 36(p+1)²  - 12(p+1)(p+9)

= 36(p² + 1 + 2p) - 12 (p²+9p+p+9)

= 36p² + 36 + 72p - 12p²- 108p-12p- 108

= 36p²-12p² +72p - 108p-12p +36- 108

= 24p²  - 48p - 72

= 24(p² - 2p - 3)

Since, roots of given equation are  equal. D= 0

0 =  24(p² - 2p - 3)

(p² - 2p - 3)= 0

p² - 3p +p - 3 = 0

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

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

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

p ≠ 0 or p = 3

Hence, required value of p is 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

(x-3)= 0  or  (x-3)= 0

x = 3 or x= 3

Hence , the roots are x =3, 3.

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