Solve the quadratic equation 4x2+4bx-(a2-b2)=0 by using the completing the square method !
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Answered by
59
Hi dear !
Completing the square method :-
divide the whole equation by coefficient of x² , which is 4 :-
x² + bx - (a² -b²)/4 = 0
now , take the square of the half of the coefficient of "x" , that is b :-
x² + bx + (b/2)² - (b/2)² - (a² -b²)/4 = 0
x² + bx + (b/2)² = (b/2)² +(a² -b²)/4
x² + bx + (b/2)² this can be written as => (x + b/2)²
(x + b/2)² = b² + a² - b²/4
(x + b/2)² = a²/4
x + b/2 = ± a/2
x = +a - b/2
x = -(a - b)/2
The roots are a - b/2 and -a - b/2
Completing the square method :-
divide the whole equation by coefficient of x² , which is 4 :-
x² + bx - (a² -b²)/4 = 0
now , take the square of the half of the coefficient of "x" , that is b :-
x² + bx + (b/2)² - (b/2)² - (a² -b²)/4 = 0
x² + bx + (b/2)² = (b/2)² +(a² -b²)/4
x² + bx + (b/2)² this can be written as => (x + b/2)²
(x + b/2)² = b² + a² - b²/4
(x + b/2)² = a²/4
x + b/2 = ± a/2
x = +a - b/2
x = -(a - b)/2
The roots are a - b/2 and -a - b/2
Answered by
3
Answer:
Hey mate first see that while doing solving any type of quadratic equation you have to stress on these points :-(I) Bringing a to the value of 1, (ii) you have to add and subtract a fixed term according to the condition of the quadratic equation, (iii) Always put + and minus sign before the root of the number.
Step-by-step explanation:
4x^2+4bx-(a^2-b^2)=0 =>4bx^2+4bx/4-(a^2-b^2/4) =0 =>x^2+bx-(a^2-b^2/4) =0 =>(X)^2+2*x*b/2+(b/2)^2-(b/2)^2-(a^2-b^2/4)=0 =>(x+b/2)^2=(b/2)^2+(a^2-b^2/4) =>(x+b/2)^2=b^2/4+a^2-b^2/4 =>(x+b/2)^2=a^2/4 =>x+b/2=+-√a^2/4=+-a/2 => x=a/2-b/2=a-b/2 or x=-(a+b)/2. Hope it helps you. Please mark me as brainliest.
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