solve the following quadratic equation( if they exist) by the method of completing square : 1) 5x^2+16x+3=0
Answers
Step-by-step explanation:
Given :-
5x²+16x+3 = 0
To find :-
Solve the following quadratic equation( if they exist) by the method of completing square ?
Solution :-
Given equation is 5x²+16x+3 = 0
On comparing with the standard quadratic equation ax²+bx+c = 0
a = 5
b = 16
c = 3
The value of the discriminant of the equation is
D = b²-4ac
=> D = 16²-4(5)(3)
=> D = 256 - 60
=> D = 196
Therefore, D > 0 then the equation has two distinct and real roots.
Given equation is 5x²+16x+3 = 0
On dividing by 5 both sides then
=> (5x²/5)+(16x/5)+(3/5) = (0/5)
=> x²+(16x/5)+(3/5) = 0
=> x²+(2/2)(16x/5)+(3/5) = 0
=> x² +(2)(x)(16/10)+(3/5) = 0
=> x² +2(x)(8/5)+(3/5) = 0
=> x²+2(x)(8/5) = 0-(3/5)
=> x²+2(x)(8/5) = -3/5
On adding (8/5)² both sides then
=> x²+2(x)(8/5)+(8/5)² = (-3/5)+(8/5)²
=> x²+2(x)(8/5)+(8/5)² = (-3/5)+(64/25)
=> [x-(8/5)]² = (-3/5)+(64/25)
=> [x-(8/5)]² = (-15+64)/25
=>[x-(8/5)]² = 49/25
=> x-(8/5) = ±√(49/25)
=> x-(8/5) = ±(7/5)
=> x = (8/5)±(7/5)
=> x = (8±7)/5
=> x = (8+7)/5 or (8-7)/5
=> x = 15/5 or 1/5
=> x = 3 or 1/5
Answer:-
The roots of the given equation are 3 and 1/5
Check:-
Given roots are 3 and 1/5
The equation is
(x-3)[x-(1/5)] = 0
=> x[x-(1/5)]-3[x-(1/5)] = 0
=> x²-(x/5)-3x+(3/5) = 0
=> (5x²-x-15x+3)/5 = 0
=> (5x²-16x+3)/5 = 0
=> 5x²-16x+3 = 0
Given equation
Verified the given relations in the given problem.
Used Method:-
- Completing the square method
Used formulae:-
- The standard quadratic equation is ax²+bx+c = 0
- The discriminant of ax²+bx+c = 0 is D = b²-4ac
- If D = b²-4ac ≥ 0 then the roots of the equation exists .
- If D> 0 the roots are distinct and real roots.
- If D<0 then the roots are no real .
- If D=0 then the roots are equal and real roots.
HOPE IT HELPS........
JAI HIND
