Question

The sum of squares of two consecutive odd no. Is 394.Find the numbers.

Answer 1

AnSwer:

• Let 1st no. be x and 2nd no. be x + 2

Here,

According to the question,

$: \implies \sf \: \: \: {x}^{2} + { (x + 2) }^{2} = 394 \\ \\ \\ : \implies \sf \: \: \: {x}^{2} + {x}^{2} + 4x + 4 = 394 \\ \\ \\ : \implies \sf \: \: \:2 {x}^{2} + 4x = 394 - 4 \\ \\ \\ : \implies \sf \: \: \:2 {x}^{2} + 4x = 390 \\ \\ \\ : \implies \sf \: \: \:2 {x}^{2} + 4x - 390 = 0 \\ \\ \\ : \implies \sf \: \: \: {x}^{2} + 2x - 195 = 0 ....(dividing \: by \: 2) \\ \\ \\ : \implies \sf \: \: \: {x}^{2} + 15x - 13x - 195 = 0...(taking \: factors) \\ \\ \\ : \implies \sf \: \: \:x(x + 15) - 13(x + 15) = 0 \\ \\ \\ : \implies \sf \: \: \:(x - 13)(x + 15) = 0 \\ \\ \\ : \implies \sf \: \: \:x = 13 \: or \: x = - 15$

Here as x can't be odd no. so,

$\implies {\underline{ \boxed{ \sf{ \purple{\:x = 13}}}}}$

Now,

$\\ \sf \: 2nd \: no. \: is :- \\ \\ \\ : \implies \sf \: \: \: x + 2 \\ \\ \\ : \implies \sf \: \: \:13 + 2 \\ \\ \\ : \implies \: \: {\underline{ \boxed{ \sf{ \purple{\:15}}}}}$

Therefore, the 2 numbers are 13 and 15.

Answer 2

$\sf{\underline{\huge{\boxed{\purple{\mathfrak{Question}}}}}}$

• The sum of squares of two consecutive odd no. Is 394.Find the numbers

$\sf{\underline{\huge{\boxed{\purple{\mathfrak{Solution}}}}}}$

• let the no. be x and x + 2

$\\ \sf\longrightarrow x ^2 + ( x + 2)^2 = 394$

$\\ \sf{\bold{\red{\boxed{(a + b)^2 = a^2 + b^2 + 2ab }}}}$

$\\ \sf\longrightarrow x^2 + x^2 + 4 + 4x = 394$

$\\ \sf\longrightarrow 2x^2 + 4 + 4x - 394 = 0$

$\\ \sf\longrightarrow 2x^2 + 4x - 390 = 0$

• taking u as common

$\\ \sf\longrightarrow 2( x^2 + 2x - 195) = 0$

$\\ \sf\longrightarrow x^2 + 2x - 195 = 0$

$\\ \sf\longrightarrow x^2 +15x - 13x - 195 = 0$

$\\ \sf\longrightarrow x ( x + 15 ) - 13 ( x + 15 ) = 0$

$\\ \sf\longrightarrow ( x + 15 ) ( x - 13 ) = 0$

⠀⠀$\begin{tabular}{|c|c|}\cline{1-2}\sf x + 15 = 0 &\sf x - 13 = 0 \\\cline{1-2}\sf x = 0 - 15 &\sf x = 0 + 13 \\\cline{1-2}\sf x = - 15 &\sf x = 13 \\\cline{1-2}\end{tabular}$

• no. cannot be negative therefore x = 13

$\sf{\underline{\boxed{\pink{\mathfrak{first \: no. = 13 }}}}}$

$\\ \sf\longrightarrow x + 2$

$\\ \sf\longrightarrow 13 + 2$

$\\ \sf\longrightarrow 15$

$\\ \sf{\underline{\boxed{\pink{\mathfrak{second\: no. = \: 15 }}}}}$

Therefore two consecutive numbers are $\sf{\underline{\boxed{\pink{\mathfrak{13 }}}}}$ and $\sf{\underline{\boxed{\pink{\mathfrak{15 }}}}}$

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