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Find the value of p for which the quadratic equation (p-1)x²-6(p+1)x+3(p+9)=0 has equal roots. Find the roots. |
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Answer» 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≠ 0Here, 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= 00 = 24(p² - 2p - 3)(p² - 2p - 3)= 0p² - 3p +p - 3 = 0p(p -3) +1(p-3)= 0(p + 1) (p +3)= 0(p + 1) (p +3)= 0p ≠ 0 or p = 3Hence, required value of p is 3Put 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)= 04x² -24x +36= 04(x² -6x +9)=0x² -6x +9 = 0x² -3x -3x +9= 0x(x -3) -3(x -3)= 0(x-3)(x-3)= 0(x-3)= 0 or (x-3)= 0x = 3 or x= 3Hence , the roots are x =3, 3.
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