(1 2a)^2 4(1 a)^2=0 ⇒ (1 2a)^2 (2 2a)^2=0 (a^2 b^2=(a b)(a+b)) (1 2a+2 2a)(1 2a (2 2a))=0 ⇒ (3 4a)(1 2a 2+2a)=0 ⇒ (3 4a)( 1)=0 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Mathematics

Properties of Square Numbers

1-2a2=41-a2

Question

$(1 - 2 a)^{2} = 4 (1 - a)^{2}$

Open in App

Solution

$(1 - 2 a)^{2} - 4 (1 - a)^{2} = 0$
$\Rightarrow (1 - 2 a)^{2} - (2 - 2 a)^{2} = 0$ $(a^{2} - b^{2} = (a - b) (a + b))$
$(1 - 2 a + 2 - 2 a) (1 - 2 a - (2 - 2 a)) = 0$
$\Rightarrow (3 - 4 a) (1 - 2 a - 2 + 2 a) = 0$
$\Rightarrow (3 - 4 a) (- 1) = 0$
$\Rightarrow 3 - 4 a = 0$
$\Rightarrow a = \frac{3}{4}$ .

Suggest Corrections

Similar questions

If $a_{1} + 2 a_{2} + 3 a_{3} = 1$ , where $a_{1}, a_{2} & a_{3}$ are all positive numbers; then what is the minimum value of $2^{a 1} + 4^{a 2} + 8^{a 3}$ ?

Q. Factorise

a^{4} - 16

is ____

Q. Simplify:

2 (a^{2} + \frac{1}{a^{2}}) - (a - \frac{1}{a}) - 7

If $a_{1}, a_{2}, a_{3}, a_{4}$ are the coefficients of any four

consecutive terms in the expansion of ( $1 + x)^{n}$ , then

$\frac{a_{1}}{a_{1} + a_{2}}$ + $\frac{a_{3}}{a_{3} + a_{4}}$ =

a^{2} b (a^{3} - a + 1) - a b (a^{4} - 2 a^{2} + 2 a) - b (a^{3} - a^{2} - 1)

Join BYJU'S Learning Program

(1−2a)2=4(1−a)2

(1−2a)2−4(1−a)2=0⇒(1−2a)2−(2−2a)2=0 (a2−b2=(a−b)(a+b))(1−2a+2−2a)(1−2a−(2−2a))=0⇒(3−4a)(1−2a−2+2a)=0⇒(3−4a)(−1)=0⇒3−4a=0⇒a=34.

$(1 - 2 a)^{2} = 4 (1 - a)^{2}$