The given quadratic equation is x2+a+1x+b=0. Since the zeroes of the given equation are 2 and –3. So, α=2 and β=-3 Now, Sum of zeroes=-Coefficient of xCoefficient of x2⇒2+-3=-a+11⇒-1=-a-1⇒a=0 Product of zeroes=Constant termCoefficient of x2⇒2×-3=b1⇒b=-6 So, a = 0 and b = −6 Hence, the correct answer is option D.
The zero of the quadratic polynomial x2+(a+1)x+b are 2 and −3, then
a) a=–7,b=–1 b) a=5,b=–1 c) a=2,b=–6 c) a=0,b=–6
If a and b are the zeroes of the quadratic polynomial f(x)=3x^2-7x-6, find a polynomial whose zeroes are (i) a^2 and b^2 (ii) 2a+3b and 3a+2b
The quadratic equations x2+ax+b=0 and x2+bx+a=0, have one common root and a≠b, then