Advertisements
Advertisements
प्रश्न
Solve for x: (x2 - 5x)2 - 7(x2 - 5x) + 6 = 0; x ∈ R.
Advertisements
उत्तर
Given equation
(x2 - 5x)2 - 7(x2 - 5x) + 6 = 0
Put x2 - 5x = y
∴ The given equation becomes
y2 - 7y + 6 = 0
⇒ y2 - 6y - y + 6 = 0
⇒ y(y - 6) -1(y - 6) = 0
⇒ y = 1, 6
But x2 - 5x = y
∴ x2 - 5x = 1
x2 - 5x - 1 = 0
Here a = 1, b = -5, c = -1
∴ x = `(-b ± sqrt(b^2 - 4ac))/(2a)`
x = `(-(-5) ± sqrt(25 + 4))/(2)`
x = `(5 ± sqrt(29))/(2)`
x2 - 5x = 6
⇒ x2 - 5x - 6 = 0
⇒ x2 - 6x + x - 6 = 0
⇒ x(x - 6) +1(x - 6) = 0
⇒ (x - 6) (x + 1) = 0
⇒ x = 6 or x = -1
Hence, the roots are -1, 6, `(5 ± sqrt(29))/(2)`.
संबंधित प्रश्न
Solve the following quadratic equation by using formula method: 5m2 + 5m – 1 = 0
Find the values of k for which the quadratic equation 9x2 - 3kx + k = 0 has equal roots.
Solve the following quadratic equation using formula method only
x2 - 4x - 1 = 0
Find the value of k so that sum of the roots of the quadratic equation is equal to the product of the roots:
kx2 + 6x - 3k = 0, k ≠ 0
If –5 is a root of the quadratic equation 2x2 + px – 15 = 0, then:
Solve for x: `5/2 x^2 + 2/5 = 1 - 2x`.
For the roots of the equation a – bx – x2 = 0; (a > 0, b > 0), which statement is true?
The roots of equation (q – r)x2 + (r – p)x + (p – q) = 0 are equal.
Prove that 2q = p + r; i.e., p, q, and r are in A.P.
The roots of quadratic equation x(x + 8) + 12 = 0 are ______.
If the roots of x2 – px + 4 = 0 are equal, the value (values) of p is ______.
