Question

$\rm(x^2+5x-2)^2+x^2+5x-7=0 \\ \\ \bf \: find \: {x}^{2} + 5x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$
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Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:x^2\:+\:5x\:=\:\dfrac{3\:\pm\:\sqrt{21}}{2}}}}$

Step-by-step-explanation:

The given equation is ( x² + 5x - 2 )² + x² + 5x - 7 = 0.

We have to find the value of x² + 5x.

Now,

( x² + 5x - 2 )² + x² + 5x - 7 = 0

By substituting ( x² + 5x ) = y, we get,

( y - 2 )² + y - 7 = 0

⇒ y² - 2 * y * 2 + 2² + y - 7 = 0 - - - [ ∵ ( a - b )² = a² - 2ab + b² ]

⇒ y² - 4y + 4 + y - 7 = 0

⇒ y² - 4y + y + 4 - 7 = 0

⇒ y² - 3y - 3 = 0

Comparing with ax² + bx + c = 0, we get,

• a = 1
• b = - 3
• c = - 3

Now,

b² - 4ac = ( - 3 )² - 4 * 1 * ( - 3 )

⇒ b² - 4ac = 9 - 4 * ( - 3 )

⇒ b² - 4ac = 9 + 12

⇒ b² - 4ac = 21

Now, we know that,

$\displaystyle{\pink{\sf\:y\:=\:\dfrac{-\:b\:\pm\:\sqrt{b^2\:-\:4ac}}{2a}}\sf\:\:\:-\:-\:-\:[\:Quadratic\:Formula\:]}$

$\displaystyle{\implies\sf\:y\:=\:\dfrac{-\:(\:-\:3\:)\:\pm\:\sqrt{21}}{2\:\times\:1}}$

$\displaystyle{\implies\boxed{\red{\sf\:y\:=\:\dfrac{3\:\pm\:\sqrt{21}}{2}}}}$

Now, by resubstituting y = ( x² + 5x ), we get,

$\displaystyle{\implies\sf\:y\:=\:x^2\:+\:5x}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:x^2\:+\:5x\:=\:\dfrac{3\:\pm\:\sqrt{21}}{2}}}}}$