Advertisements
Advertisements
प्रश्न
If x = –2 is the common solution of quadratic equations ax2 + x – 3a = 0 and x2 + bx + b = 0, then find the value of a2b.
Advertisements
उत्तर
Given quadratic equations are
ax2 + x – 3a = 0 ...(i)
x2 + bx + b = 0 ...(ii)
Since, given x = –2 is the common solution of above quadratic equation.
∴ From equations (i),
a(–2)2 + (–2) – 3a = 0
⇒ 4a – 2 – 3a = 0
⇒ a = 2
From equation (ii),
(–2)2 + b(–2) + b = 0
⇒ 4 – 2b + b = 0
⇒ b = 4
Now, a2b = (2)2 × 4 = 4 × 4 = 16
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equation by factorization method: 9x2-25 = 0
Solve the following quadratic equation for x: x2 – 2ax – (4b2 – a2) = 0
Find the roots of the following quadratic equation by factorisation:
100x2 – 20x + 1 = 0
Solve the following quadratic equation by factorization:
`sqrt(6)x^2 - 4x - 2sqrt(6) = 0`
Solve the following quadratic equations by factorization: \[\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1}, x \neq - 1, \frac{1}{3}\]
The positive value of k for which the equation x2 + kx + 64 = 0 and x2 − 8x + k = 0 will both have real roots, is
If the sum and product of the roots of the equation kx2 + 6x + 4k = 0 are real, then k =
The sum of the square of 2 consecutive odd positive integers is 290.Find them.
Find the factors of the Polynomial 3x2 - 2x - 1.
Solve the following equation by factorization
3x2= x + 4
