Advertisements
Advertisements
प्रश्न
Solve: (x – 4)(x – 2)(x- 7)(x + 1) = 16
उत्तर
(x – 4)(x – 7) (x – 2)(x + 1) = 16
⇒ (x – 4)(x – 2)(x – 7)(x + 1) = 16
⇒ (x2 – 6x + 8)(x2 – 6x – 7) = 16
Let x2 – 6x = y
(y + 8)(y – 7) = 16
⇒ y2 – 7y + 8y – 56 – 16 = 0
⇒ y2 + y – 72 = 0
⇒ (y + 9)(y – 8) = 0
y + 9 = 0
x2 – 6x + 9 = 0
(x – 3)2 = 0
x = 3, 3
or
y – 8 = 0
x2 – 6x – 8 = 0
x = `(6 +- sqrt(36 - 4(1)(- 8)))/(2(1))`
= `(6 +- sqrt(36 + 32))/2`
= `(6 + sqrt(68))/2`
= `(6 +- 2sqrt(17))/2`
= `2 ((3 +- sqrt(17)))/2`
= `3 + sqrt(17)`
Roots are `3, 3, 3 +- sqrt(17)`
APPEARS IN
संबंधित प्रश्न
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
Find all zeros of the polynomial x6 – 3x5 – 5x4 + 22x3 – 39x2 – 39x + 135, if it is known that 1 + 2i and `sqrt(3)` are two of its zeros
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 – 5)(x – 7) (x + 6)(x + 4) = 504
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