Question

Find the roots of this equation




$x {}^{2} - 4ax + 4a {}^{2} - b {}^{2}$



And the answer is 2a-b,2a+b




Please solve it soon​

Answer 1

Explanation

Given a quadratic equation $\sf x^2 - 4ax + 4a^2 - b^2$

Comparing it with the standard form Ax² + Bx +c, we get

A = 1

B = - 4a

and C = 4a² - b²

Now, we have a formula for finding the roots of a quadratic equation known as Sreedharacharya's Formula or Quadratic Formula.

According to it, the roots of the equation are :-

$\sf x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Put the respective values of a, b and c.

(Here, a, b and c are the coefficients of x², x and the constant term respectively)

So, $\sf x = \frac{-(-4a)\pm \sqrt{(-4a)^2-4(1)(4a^2 - b^2)}}{2(1)}$

$\sf\implies x = \frac{4a \pm \sqrt{16a^2 - 16a^2 + 4b^2}}{2}$

$\sf\implies x = \frac{4a \pm \sqrt{4b^2}}{2}$

$\sf\implies x = \frac{4a\pm 2b}{2}$

So, the two respective roots would be

$\sf x = \frac{4a + 2b}{2}$ and $\sf x = \frac{4a - 2b}{2}$

Cancel out 2 as common term, then we get

x = 2a + b and x = 2a - b

Answer 2

Answer:----

Step-by-step explanation:

Given Quadratic Equation :-

x² – 4ax + 4a² – b² = 0

Solution :-

Using Quadratic Formula, we will find the roots of a quadratic equation.

Putting all the values as ax² + bx +c  , we got

⇒ x² – 2 × x × 2a + (2a)² – (2a)² + 4a² – b² = 0

⇒ (x – 2a)² – b² = 0

⇒ (x – 2a)² = b²

⇒ x – 2a = ± b

⇒ x = 2a ± b

 

Hence, x =2a+ b and x = 2a-b are the two roots of the given quadratic equation.