Question

Need help with this.​

Answer 1

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Let t as y² + 5y .

${y}^{2} + 5y = t$

Rewrite the equation :

$t - \frac{36}{t} = 0$

Multiply sides by t

$t \times (t - \frac{36}{t} ) = t \times 0$

${t}^{2} - 36 = 0$

Add sides 36

${t}^{2} - 36 + 36 = 0 + 36$

${t}^{2} = 36$

$t = ± \: 6$

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So :

${y}^{2} + 5y = 6$

Subtract sides 6

${y}^{2} + 5y - 6 = 6 - 6$

${y}^{2} + 5y - 6 = 0$

$(y + 6)(y - 1) = 0$

$y = - 6 \: \: or \: \: y = 1$

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${y}^{2} +5y = - 6$

$No \: \: solution$

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Thus (( y = 1 )) and (( y = - 6 )) are the roots of the equation .

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