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Expansion of (a ± b)²

If a + b+ c...

Question

If $a + b + c = p$ and $a b + b c + c a = q$ ; find $a^{2} + b^{2} + c^{2}$ .

p^{2} - q

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p^{2} - 7 q

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p^{2} - 3 q

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p^{2} - 2 q

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Solution

The correct option is D $p^{2} - 2 q$
We know that
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
Given $a b + b c + c a = q$
$a + b + c = p$
$=> p^{2} = a^{2} + b^{2} + c^{2} + 2 (q)$
$=> a^{2} + b^{2} + c^{2} = p^{2} - 2 q$

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Similar questions

In the given figure, D is the midpoint of side BC and AE $⊥$ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that

(i) $b^{2} = p^{2} + a x + \frac{a^{2}}{4}$

(ii) $c^{2} = p^{2} - a x + \frac{a^{2}}{4}$

(iii) $(b^{2} + c^{2}) = 2 p^{2} + \frac{1}{2} a^{2}$

(iv) $(b^{2} - c^{2}) = 2 a x$

In Fig. 7.223, D is the mid-point of side BC and $A E ⊥ B C$ . If BC = a AC=b, AB=c, ED=z, AD = p and AE =h, prove that:

(i) $b^{2} = p^{2} + a x + \frac{a^{2}}{4}$
(ii) $c^{2} = p^{2} + a x + \frac{a^{2}}{4}$
(iii) $b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{4}$

\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1

\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0

\frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} + \frac{r^{2}}{c^{2}}

p^{2} + q^{2} - a^{2} - b^{2} + 2 p q + 2 a b + p + q - a + b

b^{2} = p^{2} + a x + \frac{a^{2}}{4}

c^{2} = p^{2} - a x + \frac{a^{2}}{4}

(b^{2} + c^{2}) = 2 p^{2} + \frac{1}{2} a^{2}

(b^{2} - c^{2}) = 2 a x

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