Question

solve the quadratic equation 4x2 -4a2x+(a4-b4)=0

Answer 1

4x²-4a²x+(a⁴-b⁴)=0
or, 4x²-2{(a²+b²)+(a²-b²)}x+(a²+b²)(a²-b²)=0
or, 4x²-2(a²+b²)x-2(a²-b²)x+(a²+b²)(a²-b²)=0
or, 2x{2x-(a²+b²)}-(a²-b²){2x-(a²+b²)}=0
or, {2x-(a²+b²)}{2x-(a²-b²)}=0
Either, 2x-(a²+b²)=0
or, 2x=a²+b²
or, x=(a²+b²)/2
Or, 2x-(a²-b²)=0
or, 2x=a²-b²
or, x=(a²-b²)/2
∴, x=(a²+b²)/2, (a²-b²)/2 Ans..

Answer 2

The solution of the equation are $x=\frac{a^2+ b^2}{2},\frac{a^2- b^2}{2}$.

Step-by-step explanation:

Given : The quadratic equation $4x^2-4a^2x+(a^4-b^4)=0$

To find : Solve the quadratic equation ?

Solution :

Using quadratic formula of equation $ax^2+bx+c=0$ is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, $a=4,\ b=-4a^2,\ c=a^4-b^4$

Substitute the value,

$x=\frac{-(-4a^2)\pm\sqrt{(-4a^2)^2-4(4)(a^4-b^4)}}{2(4)}$

$x=\frac{4a^2\pm\sqrt{(4b^2)^2}}{8}$

$x=\frac{4a^2\pm 4b^2}{8}$

$x=\frac{a^2\pm b^2}{2}$

$x=\frac{a^2+ b^2}{2},\frac{a^2- b^2}{2}$

Therefore, the solution of the equation are $x=\frac{a^2+ b^2}{2},\frac{a^2- b^2}{2}$.

#Learn more

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