Question

А) ( у + 4 )2 = 3у2 - 8;

б) (х - 3 )2 =2 ( х + 1 ).

3. а) 14 х2- 17 х + 3 = 0;

б) ( 9 - 4х )2 =5 ( 4х + 1)​

Answer 1

a)

$( y + 4 )^2 = 3y^2 - 8\\y^2+8y+16-3y^2+8=0\\-2y^2+8y+24=0\\y^2-4y-12=0\\$

По т.Виета

$x_1=6\\x_2=-2$

б)

$(x - 3 )^2 =2 ( x + 1 )\\x^2-6x+9-2x-2=0\\x^2-8x+7=0$

По т.Виета

$x_1=7\\x_2=1$

а)

$14 x^2- 17 x + 3 = 0\\D=(-17)^2-4*3*14=121=11^2\\\\x_1=\frac{17+11}{2*14} =1\\\\x_2=\frac{17-11}{2*14}= \frac{6}{28}=\frac{3}{14}$

б)

$( 9 - 4x )^2 =5 ( 4x + 1)\\81-72x+16x^2-20x-5=0\\16x^2-92x+76=0\\4x^2-23x+19=0\\D=(-23)^2-4*4*19=225=15^2\\\\x_1=\frac{23+15}{2*4}= 4,75\\\\x_2=\frac{23-15}{2*4} =1$