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Find the values of the following expressions:
(i) $16x^2 + 24x + 9$ when $x = \frac{7}{4}$
(ii) $64x^2 + 81y^2 + 144xy$ when $x = 11$ and $y = \frac{4}{3}$
(iii) $81x^2 + 16y^2 - 72xy$ when $x = \frac{2}{3}$ and $y = \frac{3}{4}$
To do:
We have to find the values of the given expressions.
Solution:
Here, we have to find the values of the given expressions. So, simplifying the given expressions using the identities $(a+b)^2=a^2+2ab+b^2$.............(I) and $(a-b)^2=a^2-2ab+b^2$.............(II) and substituting the values of $x$ and $y$, we can find the required values.
(i) The given expression is $16x^2 + 24x + 9$.
$16x^2 + 24x + 9=(4x)^2+2\times 4x \times3+(3)^2$ [$24x=2\times 4x \times3$]
$16x^2 + 24x + 9=(4x+3)^2$ (Using (I), $a=4x$ and $b=3$)
Substituting $x = \frac{7}{4}$ in $(4x+3)^2$, we get, $(4x+3)^2=[4\times\frac{7}{4}+3]^2$ $(4x+3)^2=(7+3)^2$ $(4x+3)^2=(10)^2$ $(4x+3)^2=100$
The value of the expression $16x^2 + 24x + 9$ when $x = \frac{7}{4}$ is $100$.
(ii) The given expression is $64x^2 + 81y^2 + 144xy$
$64x^2 + 81y^2 + 144xy=(8x)^2+2\times 8x \times9y+(9y)^2$ [$144xy=2\times 8x \times9y$]
$64x^2 + 81y^2 + 144xy=(8x+9y)^2$ (Using (I), $a=8x$ and $b=9y$)
Substituting $x = 11$ and $y = \frac{4}{3}$ in $(8x+9y)^2$, we get,
$(8x+9y)^2=[8\times11+9\times\frac{4}{3}]^2$
$(8x+9y)^2=(88+3\times4)^2$
$(8x+9y)^2=(88+12)^2$
$(8x+9y)^2=(100)^2$
$(8x+9y)^2=10000$
The value of the expression $64x^2 + 81y^2 + 144xy$ when $x = 11$ and $y = \frac{4}{3}$ is $10000$.
(iii) The given expression is $81x^2 + 16y^2 - 72xy$
$81x^2 + 16y^2 - 72xy=(9x)^2-2\times 9x \times 4y+(4y)^2$ [$72xy=2\times 9x \times 4y$]
$81x^2 + 16y^2 - 72xy=(9x-4y)^2$ (Using (II), $a=9x$ and $b=4y$)
Substituting $x = \frac{2}{3}$ and $y = \frac{3}{4}$ in $(9x-4y)^2$, we get,
$(9x-4y)^2=[9\times\frac{2}{3}-4\times\frac{3}{4}]^2$
$(9x-4y)^2=(3\times2-1\times3)^2$
$(9x-4y)^2=(6-3)^2$
$(9x-4y)^2=(3)^2$
$(9x-4y)^2=9$
The value of the expression $81x^2 + 16y^2 - 72xy$ when $x = \frac{2}{3}$ and $y = \frac{3}{4}$ is $9$.
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