Question

simplify (2√5+3√2)² +(2√5-3√2)²​

Answer 1

Answer:

$(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2=76$

Step-by-step explanation:

Given : Expression $(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2$

To find : Simplify the expression ?

Solution :

Expression $(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2$

Open the squaring term by identity, $(a+b)^2=a^2+b^2+2ab$

$=(2\sqrt5)^2+(3\sqrt2)^2+2(2\sqrt5)(3\sqrt2)+(2\sqrt5)^2+(3\sqrt2)^2-2(2\sqrt5)(3\sqrt2)$

Cancel the like terms,

$=20+18+20+18$

$=40+36$

$=76$

Therefore, $(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2=76$

Answer 2

The value of the $(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2$ is equal to 76.

Step-by-step explanation:

We have,

$(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2$

To find, the value of the $(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2$ = ?

∴ $(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2$

Here, a = $2\sqrt5$ and b = $3\sqrt2$

We know that,

The algebraic identity,

$(a+b)^2 +(a-b)^2=2(a^2+b^2)$

∴ $(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2$

$=2[(2\sqrt5)^2+(3\sqrt2)^2]$

= 2[4(5) + 9(2)]

= 2(20 + 18)

= 2(38)

= 76

∴ The value of the $(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2$ = 76

Thus, the value of the $(2\sqrt5+3\sqrt2)^2+(2\sqrt5-3\sqrt2)^2$ is equal to 76.