If a and b are the roots of the quadratic equation x²+px+12=0 with the condition a-b=1, then what is the value of p?
Answers
Answer:
Sum of roots = a + b = -p.
Product of roots = ab = 12.
(a - b)^2 = (a + b)^2 - 4(ab) = p^2 - 48 = 1. ==>*** p = ±7.***
Case 1: x^2 + 7x + 12 = (x+4)(x+3)=0. x = -4 or -3.
Case 2: x^2 - 7x + 12 = (x -4)(x-3)=0. x = 4 or 3
Answer:
For a quadratic equation of the form: dx2+ex+f,
The sum of the roots =−e/d
The product of the roots =f/d
In the equation : x2+px+12
d= coefficient of x2=1
e = coefficient of x=p
f = constant = 12.
1.Sum of the roots: a+b=−e/d=−p/1=−p
2. Product of the roots: ab=f/d=12
We have a−b=1.
a=1+b.
Substituting a= 1+b in Equation 2: ab =12
(1+b)b = 12
b2+b=12
b2+b−12=0
(b+4)(b-3) = 0
Thus b = - 4 or b = 3
If b = -4 then a = -3, since a-b = 1, a = 1+b
If b = 3 then a = 4 since a-b = 1, a = 1+b
From equation 1: Sum of roots: a+b = -p,
Case 1: If a=-3 and b= -4, from equation 1, we have:
a+b = -p = -7, thus p = 7
Case 2: If a= 4 and b= 3, from equatiob 1, we have:
a+b = -p = 7, thus a+b = -p = -7, thus p = -7
The two values of p are: 7, -7.