Question

please help in this question​

Answer 1

Answer:

x = - 6 + - \sqrt{6 } - 6 + \sqrt{31 }x=−6+−

6

−6+

31

Answer 2

Step-by-step explanation:

Given:

${( {x}^{2} + 12x)}^{2} + 35( {x}^{2} + 12x) + 150 = 0$

To find: Solve the equation.

Solution:

Let

$\bold{{x}^{2} + 12x = y}$

Put this to given equation

$\bold{\green{ {y}^{2} + 35y + 150 = 0 }}$

Solve this quadratic equations in y

${y}^{2} + 30y + 5y + 150 = 0 \\ \\ y(y + 30) + 5(y + 30) = 0 \\ \\ (y + 30) (y + 5) = 0 \\ \\ y = - 30 \\ \\ or \\ \\ y = - 5$

To find value of x:

1. put y=-30 in eq1

${x}^{2} + 12x = - 30 \\ \\ {x}^{2} + 12x + 30 = 0$

To find value of x, use quadratic formula

$\boxed{\bold{x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a}}}$

$a = 1 \\ b = 12 \\ c = 30$

$x_{1,2} = \frac{ - 12 ± \sqrt{144 - 120} }{2} \\ \\ x_{1,2} = \frac{ - 12 ± \sqrt{24} }{2} \\ \\ x_{1,2} = \frac{ - 12 ± 2 \sqrt{6} }{2} \\ \\ x_{1,2} = - 6 ± \sqrt{6} \\ \\ x_1=- 6 + \sqrt{6}\\\\x_2=- 6 - \sqrt{6}$

2) Put y=-5 in eq1

${x}^{2} + 12x = - 5 \\ \\ {x}^{2} + 12x + 5 = 0$

$a = 1 \\ b = 12 \\ c = 5$

$x_{3,4} = \frac{ - 12 ± \sqrt{144 - 20} }{2} \\ \\ x_{3,4} = \frac{ - 12 + \sqrt{124} }{2} \\ \\ x_{3,4} = \frac{ - 12 + 2 \sqrt{31} }{2} \\ \\ x_3 = - 6 + \sqrt{31} \\ \\ x_4 = - 6 - \sqrt{31}$

Final answer:

Solution of equation are

$\bold{\red{x_1 = - 6 + \sqrt{6} }} \\ \bold{\green{x_2 = - 6 - \sqrt{6}}} \\ \bold{\pink{x_3= - 6 + \sqrt{31}}} \\\bold{\orange{ x_4 = - 6 - \sqrt{31}}}$

Hope it helps you.

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