Question

решите уравнение методом выделения квадрата двучлена 

Answer 1

a)

x^2-6x+8=0

x^2-6x+8+1-1=0

x^2-6x+9=1

(x-3)^2=1

x-3=1

x-3=-1

x₁=4

x₂=2

б)

x^2+10x+9=0

x^2+10x+9+16-16=0

x^2+10x+25=16

(x+5)^2=16

x+5=4

x+5=-4

x₁=-1

x₂=-9

в)

x^2-x-2=0

x^2-x+0.25-2.25=0

(x-0.5)^2=2.25

x-0.5=1.5

x-0.5=-1.5

x₁=2

x₂=-1

г)

x^2+3x-40=0

x^2+3x+2.25-42.25=0

x^2+3x+2.25=42.25

(x+1.5)^2=42.25

x+1.5=6.5

x+1.5=-6.5

x₁=5

x₂=-8

Answer 2

a) $x^{2}-6x+8=0\x^{2}-2cdot{3}x+3^{2}-1=0\(x-3)^{2}-1=0\(x-3-1)(x-3+1)=0\(x-4)(x-2)=0\x-4=0;x-2=0\x_{1}=4;x_{2}=2$

 

б)$x^{2}+10x+9=0\x^{2}+2cdot5x+5^{2}-16=0\(x+5)^{2}-4^{2}=0\(x+5-4)(x+5+4)=0\(x+1)(x+9)=0\x+1=0;x+9=0\x_{1}=-1;x_{2}=-9$

 

в)$x^{2}-x-2=0\x^{2}-2cdot(frac{1}{2})x+frac{1}{4}-2frac{1}{4}=0\(x-frac{1}{2})^{2}-(frac{3}{2})^{2}=0\(x-frac{1}{2}-frac{3}{2})(x-frac{1}{2}+frac{3}{2})=0\(x-2)(x+1)=0\x-2=0;x+1=0\x_{1}=2;x_{2}=-1$

 

г)$x^{2}+3x-40=0\x^{2}+2cdotfrac{3}{2}x+frac{9}{4}-40frac{9}{4}=0\(x+frac{3}{2})^{2}-frac{169}{4}=0\(x+frac{3}{2}-frac{13}{2})(x+frac{3}{2}+frac{13}{2})=0\(x-5)(x+8)=0\x_{1}=5;x_{2}=-8$