4. Determine the value of m for which the
equation (4-m) x2 + (2m+4)x+(8m+1) =0
may have equal roots.
Answers
Step-by-step explanation:
Given :-
(4-m)x²+(2m+4)x+(8m+1) = 0
To find :-
Determine the value of m for which the
equation (4-m) x2 + (2m+4)x+(8m+1) =0
may have equal roots.?
Solution :-
Given Quadratic equation is
(4-m)x²+(2m+4)x+(8m+1) = 0
On Comparing this with the standard quadratic equation ax²+bx+c = 0
We have,
a = 4-m
b = 2m+4
c = 8m+1
We know that
If a quadratic equation ax²+bx+c = 0 has equal roots then the discriminant (D)=b²-4ac = 0.
Now,
The discriminant (D)=b²-4ac = 0.
=>(2m+4)²-4(4-m)(8m+1) = 0
=> (2m)²+2(2m)(4)+(4)²-4(4-m)(8m+1) = 0
Since (a+b)² = a²+2ab+b²
=> 4m²+16m+16-4(32m+4-8m²-m) = 0
=> 4m²+16m+16-4(-8m²+31m+4) = 0
=>4(m²+4m+4)-4(-8m²+31m+4) = 0
=> 4[(m²+4m+4)-(-8m²+31m+4) ]=0
=> (m²+4m+4)-(-8m²+31m+4) = 0/4
=> (m²+4m+4)-(-8m²+31m+4) = 0
=> m²+4m+4+8m²-31m-4 = 0
=> (m²+8m²)+(4m-31m)+(4-4) = 0
=> 9m²-27m+0 = 0
=> 9m²-27m = 0
=> 9m (m-3) = 0
=> 9m = 0 or m-3 = 0
=> m = 0/9 or m= 0+3
=> m = 0 or m = 3
Therefore, m = 0 or 3
Answer:-
The value of m for the given problem is 0 and 3
Check:-
If m = 3 then (4-m)x²+(2m+4)x+(8m+1) = 0
=> (4-3)x²+(2(3)+4)x+(8(3)+1) = 0
=> x²+10x+25 = 0
=> x²+2(x)(5)+5² = 0
=> (x+5)² = 0
=> (x+5)(x+5) = 0
=> x = -5 and -5
And
If x = 0 then (4-m)x²+(2m+4)x+(8m+1) = 0
=> (4-0)x²+(2(0)+4)x+(8(0)+1) = 0
=> 4x²+4x+1 = 0
=> (2x)²+2(2x)(1)+1² = 0
=> (2x+1)² = 0
=> (2x+1)(2x+1) = 0
=> 2x+1 = 0
=> x = -1/2 and -1/2
Verified the given relations in the given problem.
Used formulae:-
- (a+b)² = a²+2ab+b²
- The standard quadratic equation is ax²+bx+c = 0
- If a quadratic equation ax²+bx+c = 0 has equal roots then the discriminant (D)=b²-4ac = 0.
Points to know:-
- If (D)=b²-4ac < 0 then it has no real roots.
- If (D)=b²-4ac > 0 then it has two distinct real roots.