Question

помогите пожалуйста❤
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Answer 1

1)

1.

$4x^{2}-12=0\\\\4x^{2}=12\\\\x^{2}=3\\\\x_{1}=\sqrt{3} \\\\x_{2}=-\sqrt{3}$

2.

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

3.

$15x^{2} -4x-3=0\\\\D=16+180=196\\\\x_{1}=\frac{4+14}{30} =\frac{18}{30}=\frac{3}{5} \\\\x_{1}=\frac{4-14}{30}=-\frac{1}{3}$

4.

$x^{2} -7x+4=0\\\\D=49-16=33\\\\x_{1}=\frac{7+\sqrt{33} }{2} \\\\x_{2}=\frac{7-\sqrt{33} }{2}$

5. корней нет, так как дискриминант отрицательный

$x^{2} +5x+9=0\\\\D=25-32=-7$

6.

$(2x+3)(2x-3)-(x-2)(x+1)=7\\\\4x^{2}-9-(x^{2}-x-2)=7\\\\4x^{2}-9-x^{2}+x+2=7\\\\3x^{2}+x-14=0\\\\D=1+168=169\\\\x_{1}=\frac{-1+13}{6} =2\\\\x_{1}=\frac{-1-13}{6}=-\frac{7}{3}$

2)

$x^{2}-x-42=0\\\\D=1+168=169\\\\x_{1}=\frac{1+13}{2} =7\\\\x_{2}=\frac{1-13}{2}=-6\\\\ x^{2}-x-42=(x-7)(x+6)$

3)

1.

$x-10x^{2}+9=0\\\\10x^{2}-x-9=0\\\\D=1+360=361\\\\x_{1}=\frac{1+19}{20} =1\\\\x_{2}=\frac{1-19}{20}=-\frac{9}{10}$

2.

$2(x+3)^{2}-5(x+3)+2=0$

Заменим $t=x+3$

$2t^{2}-5t+2=0\\\\D=25-16=9\\\\t_{1}=\frac{5+3}{4} =2\\\\t_{1}=\frac{5-3}{4}=0.5$

$x+3=2\\\\x=-1\\\\x+3=0.5\\\\x=-2.5$