Question

a2x2-abx-2b2=0  fined the root by completing square

Answer 1

See the pic .........

Answer 2

The roots of the equation $\bf \red{{a}^{2} {x}^{2} - abx - 2 {b}^{2} = 0}$are $\bf \frac{2b}{a}$ and $\bf \frac{- b}{a}$.

Given:

• ${a}^{2} {x}^{2} - abx - 2 {b}^{2} = 0$

To find:

• Find the roots by completing the square method.

Solution:

Identity to be used:

1. $\bf ( {x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}$
2. $\bf {x}^{2} - {y}^{2} = (x + y)(x - y)$

Step 1:

Complete the square in LHS.

Take 2b² to the RHS.

${a}^{2} {x}^{2} - abx = 2 {b}^{2}$

To complete the whole square, Add b²/4 both sides.

${(ax)}^{2}- 2(ax) .\frac{b}{2} + \left( { \frac{b}{2} } \right)^{2} = 2 {b}^{2} + \left( { \frac{b}{2} } \right)^{2}$

$\bf \left({ ax - \frac{b}{2} } \right)^{2} = 2 {b}^{2} + { \frac{ {b}^{2} }{4} }$

Step 2:

Simplify the RHS.

$\left({ ax - \frac{b}{2} } \right)^{2} = { \frac{ 8 {b}^{2} + {b}^{2} }{4} }$

$\left({ ax - \frac{b}{2} } \right)^{2} = { \frac{ 9{b}^{2} }{4} }$

$\left({ ax - \frac{b}{2} } \right)^{2} = \left({ \frac{ 3{b} }{2} } \right)^{2}$

$\left({ ax - \frac{b}{2} } \right)^{2} - \left({ \frac{ 3{b} }{2} } \right)^{2} = 0$

It is similar to the identity 2.

Simplify this

$\left(ax - \frac{b}{2} - \frac{3b}{2} \right)\left(ax - \frac{b}{2} + \frac{3b}{2} \right) = 0$

$\bf \left(ax - 2b \right)\left(ax + b \right) = 0$

Step 3:

Find the roots of the equation.

$ax - 2b = 0$

$\bf x = \frac{2b}{a}$

or

$ax + b = 0$

$\bf x = \frac{ - b}{a}$

Thus,

Roots of equation are 2b/a and -b/a.

Learn more:

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