If $a = 6 p$ and $b = 3 p$ , then $(a + b)^{2} + (a - b)^{2}$ = .

72 p^{2} - 18 q^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

72 p^{2} + 18 q^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

18 p^{2} + 72 q^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $72 p^{2} + 18 q^{2}$

Given: $a = 6 p$ and $b = 3 q$
To find: $(a + b)^{2} + (a - b)^{2}$

We know that,
$(a + b)^{2} = a^{2} + 2 a b + b^{2} (a - b)^{2} = a^{2} - 2 a b + b^{2}$

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

On substituting the values of $a$ and $b$ , we get,
$(a + b)^{2} + (a - b)^{2} = (6 p + 3 q)^{2} + (6 p - 3 q)^{2} = 2 (6 p)^{2} + 2 (3 q)^{2} = 2 (36 p^{2}) + 2 (9 q)^{2} = 72 p^{2} + 18 q^{2}$

Suggest Corrections

Similar questions

If a = 6 p and b = 3 p,

then find the expansion of

(a + b)^{2} + (a - b)^{2} .

If a = 6 p and b = 3 p,

then find the expansion of

(a + b)^{2} + (a - b)^{2} .

a = 6 p

b = 3 p

(a + b)^{2} + (a - b)^{2}

\frac{a}{b}

\frac{(a sinθ - b cosθ)}{(a \sin θ + b cosθ)}

\frac{(a^{2} + b^{2})}{(a^{2} - b^{2})}

\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}

\frac{a^{2}}{(a^{2} + b^{2})}

\frac{b^{2}}{(a^{2} + b^{2})}

If sin $θ$ = $\frac{a}{b}$ then cos $θ$ = ?

(a) $\frac{b}{\sqrt{b^{2} - a^{2}}}$ (b) $\frac{\sqrt{b^{2} - a^{2}}}{b}$ (c) $\frac{a}{\sqrt{b^{2} - a^{2}}}$ (d) $\frac{b}{a}$

Join BYJU'S Learning Program