If the difference between the roots of the equation x^2+ ax +8 = 0 is 2 find the value of a..
If f(x)= ax^2+bx+c then sum of roots =-b/a and product of roots=c/a
So let the roots be p and q
p+q=-a
p-q=2
pq=8
(p-q)^2=(p+q)^2-4pq
By substituting the values you'll get a=+6 or -6
So let the roots be p and q
p+q=-a
p-q=2
pq=8
(p-q)^2=(p+q)^2-4pq
By substituting the values you'll get a=+6 or -6
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let P and Q are the roots of x²+ ax + 8 =0
P+Q = -a
PQ = 8
question said that
P - Q = 2
take square both sides
(P - Q)² = 4
(P + Q)² -4PQ = 4
(-a)² -4(8) = 4
a² -32 = 4
a² = 36
a = ±6
a = 6 and -6
P+Q = -a
PQ = 8
question said that
P - Q = 2
take square both sides
(P - Q)² = 4
(P + Q)² -4PQ = 4
(-a)² -4(8) = 4
a² -32 = 4
a² = 36
a = ±6
a = 6 and -6
Answer is +8&-8
Plzz do & give right answer
no this is correct answer
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So let the roots be p and q
p+q=-a
p-a=2
pq=8
(p-q)^2=(p+q)^2-4pq
By substituting the values you'll get a=+6 or -6