Advertisements
Advertisements
प्रश्न
Solve the following by reducing them to quadratic form:
`sqrt(x^2 - 16) - (x - 4) = sqrt(x^2 - 5x + 4)`.
Advertisements
उत्तर
Given equation
`sqrt(x^2 - 16) - (x - 4) = sqrt(x^2 - 5x + 4)`
⇒ `sqrt((x + 4) (x - 4)) - (x - 4) = sqrt((x - 1) (x - 4)`
⇒ `sqrt(x - 4) [sqrt(x + 4) - sqrt(x - 4) - sqrt(x - 1)] = 0`
⇒ Either, `sqrt(x - 4) = 0`
⇒ x - 4 = 0
⇒ x = 4 ...(By squaring on both sides)
or
`sqrt(x + 4) - sqrt(x - 4) - sqrt(x - 1) = 0`
⇒ `sqrt(x + 4) - sqrt(x - 4) = sqrt(x - 1)`
Squaring both sides we get
`x + 4 + x - 4 - 2sqrt((x + 4) (x - 4)) = x - 1`
⇒ `2x - 2sqrt(x^2 - 16)) = x - 1`
⇒ `-2sqrt(x^2 - 16) = x - 2x -1 = -x -1`
= -(x + 1)
⇒ `2sqrt(x^2 - 16)) = x + 1`
Squaring again, 4(x2 - 16) = x2 + 2x + 1
⇒ 4x2 - 64 - x2 - 2x - 1 = 0
⇒ 3x2 - 2x - 65 = 0
⇒ 3x2 - 15x + 13x - 65 = 0
⇒ 3x(x - 5) + 13(x - 5) = 0
⇒ (x - 5) + (3x + 13) = 0
⇒ x - 5 = 0 or 3x + 13 = 0
⇒ x = 5 or x = `(-13)/(3)`
x = 5.
Hence, the solutions are 4, 5.
संबंधित प्रश्न
Solve the following quadratic equations by factorization:
`(x+3)/(x-2)-(1-x)/x=17/4`
The sum of two natural numbers is 15 and the sum of their reciprocals is `3/10`. Find the numbers.
Divide 57 into two parts whose product is 680.
Write the set of value of 'a' for which the equation x2 + ax − 1 = 0 has real roots.
The sum of the square of 2 consecutive odd positive integers is 290.Find them.
Solve equation using factorisation method:
2(x2 – 6) = 3(x – 4)
Solve equation using factorisation method:
2x2 – 9x + 10 = 0, when:
- x ∈ N
- x ∈ Q
Solve the following quadratic equation by factorisation method:
`(x + 3)/(x - 2) - (1 - x)/x = (17)/(4)`.
Solve the following equation by factorisation :
x(x + 1) + (x + 2)(x + 3) = 42
Find the roots of the following quadratic equation by the factorisation method:
`2/5x^2 - x - 3/5 = 0`
