9x²-9(p+q)x +(2p²+5pq+2q²)=0 solve for x
Answers
Answered by
8
9x²-9(p+q)x+(2p²+5pq+2q²)=0 by using quadratic formula
Sol:
9x2 - 9(a + b)x + (2a2 + 5ab + 2b2) = 0
a = 9, b = - 9(a + b), c = (2a2 + 5ab + 2b2)
Roots of the equation ax2 + bx + c are [-b ± (√b2 - 4ac)/2a]
= [9(a + b) ± √[81(a + b)2 - 36(2a2 + 5ab + 2b2)]/18]
= 9a + 9b ± [√9a2 + 9b2 - 18ab] / 18
= 9a + 9b ± (a - b)/6
= (55 a + 53b) / 6, (53a + 55b) / 6
Hence, the roots of the equation are (55 a + 53b) / 6, (53a + 55b) / 6.
Sol:
9x2 - 9(a + b)x + (2a2 + 5ab + 2b2) = 0
a = 9, b = - 9(a + b), c = (2a2 + 5ab + 2b2)
Roots of the equation ax2 + bx + c are [-b ± (√b2 - 4ac)/2a]
= [9(a + b) ± √[81(a + b)2 - 36(2a2 + 5ab + 2b2)]/18]
= 9a + 9b ± [√9a2 + 9b2 - 18ab] / 18
= 9a + 9b ± (a - b)/6
= (55 a + 53b) / 6, (53a + 55b) / 6
Hence, the roots of the equation are (55 a + 53b) / 6, (53a + 55b) / 6.
Similar questions