Advertisements
Advertisements
Question
Solve: (x – 5)(x – 7) (x + 6)(x + 4) = 504
Solution
(x – 5)(x + 4)(x – 7)(x + 6) = 504
(x2 – x – 20)(x2 – x – 42) = 504
Let y = (x2 – x)
(y – 20)(y – 42) = 504
⇒ y2 – 42y – 20y + 840 = 504
⇒ y2 – 62y + 336 = 0
⇒ (y – 56)(y – 6) = 0
⇒ (y – 56) = 0 or (y – 6) = 0
⇒ x2 – x – 56 = 0 or x2 – x – 6 = 0
⇒ (x – 8)(x + 7) = 0 or (x – 3)(x + 2) = 0
⇒ x = 8, – 7 or x = 3, – 2
The roots are 8, – 7, 3, – 2
APPEARS IN
RELATED QUESTIONS
Solve the cubic equation: 2x3 – x2 – 18x + 9 = 0 if sum of two of its roots vanishes
Solve the equation 9x3 – 36x2 + 44x – 16 = 0 if the roots form an arithmetic progression
Solve the equation 3x3 – 26x2 + 52x – 24 = 0 if its roots form a geometric progression
Determine k and solve the equation 2x3 – 6x2 + 3x + k = 0 if one of its roots is twice the sum of the other two roots
Solve the cubic equations:
2x3 – 9x2 + 10x = 3
Solve the cubic equations:
8x3 – 2x2 – 7x + 3 = 0
Solve the equation:
x4 – 14x2 + 45 = 0
Solve: (x – 4)(x – 2)(x- 7)(x + 1) = 16
Solve: (2x – 1)(x + 3)(x – 2)(2x + 3) + 20 = 0
Choose the correct alternative:
A zero of x3 + 64 is
Choose the correct alternative:
If α, β and γ are the zeros of x3 + px2 + qx + r, then `sum 1/alpha` is
Choose the correct alternative:
The polynomial x3 – kx2 + 9x has three real roots if and only if, k satisfies
Choose the correct alternative:
If x3 + 12x2 + 10ax + 1999 definitely has a positive zero, if and only if
Choose the correct alternative:
The polynomial x3 + 2x + 3 has
Choose the correct alternative:
The number of positive roots of the polynomials `sum_("j" = 0)^"n" ""^"n""C"_"r" (- 1)^"r" x^"r"` is
