If $a - 2 b = 15$ $a - 2b = 15$ and $a b = 11$ $ab = 11$ then find the value of $a^{2} + 4 b^{2}$ $a^2 + 4b^2$ ?

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Answer 1 · 2016-06-23T18:34:56

$a^{2} + 4 b^{2} = 269$ $a^2+4b^2=269$

${(a - 2 b)}^{2} = a^{2} - 4 a b + 4 b^{2} = 15^{2}$ $(a-2b)^2=a^2-4ab+4b^2= 15^2$

then

$a^{2} + 4 b^{2} = 15^{2} + 4 a b = 15^{2} + 4 \times 11 = 269$ $a^2+4b^2=15^2+4 ab = 15^2+4 xx 11 =269$

Answer 2 · 2016-06-23T18:49:22

$a^{2} + 4 b^{2} = 269 .$ $a^2+4b^2=269.$

Method I

Given that, $a - 2 b = 15$ $a-2b=15$
$:. (a-2b)^2=15^2=225.$
$:. a^2-2*a*2b+4b^2=225,$ i.e., $a^2-4ab+4b^2=225.$

Letting, $ab=11$ in this, we have, $a^2-4(11)+4b^2=225.$

Hence, $a^2+4b^2=225+44=269.$

Method II

Notice that $(a+2b)^2+(a-2b)^2=2(a^2+4b^2).$

Here, we replace $(a+2b)^2$ by $(a-2b)^2+8ab,$ to get,

$(a-2b)^2+8ab+(a-2b)^2=2(a^2+4b^2),.$ & now putting the given values,

$225+8*11+225=2(a^2+4b^2)=538,$ so, $(a^2+4b^2)=538/2=269,$ as before!

If a - 2b = 15a−2b=15a - 2b = 15 and ab = 11ab=11ab = 11 then find the value of a^2 + 4b^2a2+4b2a^2 + 4b^2?