Advertisements
Advertisements
प्रश्न
ax2 + (4a2 - 3b)x - 12 ab = 0
Advertisements
उत्तर
ax2 + (4a2 - 3b)x - 12 ab = 0
⇒ ax2 + 4a2x - 3bx - 12ab = 0
⇒ ax (x + 4a) - 3b(x + 4a) = 0
⇒ (ax - 3b) (x + 4a) = 0
⇒ x = `(3b)/a or -4a` are two roots of equation.
संबंधित प्रश्न
Form the quadratic equation if its roots are –3 and 4.
If (k – 3), (2k + l) and (4k + 3) are three consecutive terms of an A.P., find the value of k.
Find the value of k for which equation 4x2 + 8x – k = 0 has real roots.
If one root of the equation 2x² – px + 4 = 0 is 2, find the other root. Also find the value of p.
Find the value (s) of k for which each of the following quadratic equation has equal roots : kx2 – 4x – 5 = 0
If roots of a quadratic equation 3y2 + ky + 12 = 0 are real and equal, then find the value of ‘k’
If one root of the quadratic equation 2x2 + kx – 6 = 0 is 2, the value of k is:
Which of the following equations has two distinct real roots?
If x = 3 is one of the roots of the quadratic equation x2 – 2kx – 6 = 0, then the value of k is ______.
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.
