Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 538

а) 8x² − 14x + 5 = 0 8x² − 2 ⋅ 7x + 5 = 0 D = k² − ac = (−7)² − 8 ⋅ 5 = 49 − 40 = 9 > 0, уравнение имеет 2 корня
x₁ = $$ =\frac{-k+\sqrt{D}}{a}=\frac{7+\sqrt{9}}{8} $$ = 7 + 3/8 = 10/8 = 1,25
x₂ = $$ \frac{-k-\sqrt{D}}{a}=\frac{7-\sqrt{9}}{8} $$ = 7 − 3/8 = 4/8 = 0,5
x₁ = 1,25, x₂ = 0,5
б) 12t² + 16t − 3 = 0 12t² + 2 ⋅ 8t − 3 = 0 D = k² − ac = 8² − 12 ⋅ (−3) = 64 + 36 = 100 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-k+\sqrt{D}}{a}=\frac{8+\sqrt{100}}{12} $$ = 8 + 10/12 = 18/12 = 3/2 = 1,5
x₂ = $$ \frac{-k-\sqrt{D}}{a}=\frac{8-\sqrt{100}}{12} $$ = 8 − 10/12 = −2/12 = −1/6
x₁ = 1,5, x₂ = −1/6
в) 4p² + 4p + 1 = 0 4p² + 2 ⋅ 2p + 1 = 0 D = k² − ac = 2² − 4 ⋅ 1 = 4 − 4 = 0 = 0, уравнение имеет 1 корень
x₁ = $$ \frac{-k+\sqrt{D}}{a}=\frac{-2+\sqrt{0}}{4} $$ = −2 + 0/4 = −2/4 = −0,5
x₁ = −0,5
г) x² − 8x − 84 = 0 x² − 2 ⋅ 4x − 84 = 0 D = k² − ac = (−4)² − 1 ⋅ (−84) = 16 + 84 = 100 > 0, уравнение имеет 2 корня
x₁ = $$ =\frac{-k+\sqrt{D}}{a}=\frac{4+\sqrt{100}}{1} $$ = 4 + 10/1 = 14/1 = 14
x₂ = $$ \frac{-k-\sqrt{D}}{a}=\frac{4-\sqrt{100}}{1} $$ = 4 − 10/1 = −6/1 = −6
x₁ = 14, x₂ = −6
д) m² + 6m − 19 = 0 m² + 2 ⋅ 3m − 19 = 0 D = k² - ac = 3² − 1 ⋅ (−19) = 9 + 19 = 28 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-k+\sqrt{D}}{a}=\frac{-3+\sqrt{28}}{1} $$ = −3 + $$\sqrt{28}$$
x₂ = $$ \frac{-k-\sqrt{D}}{a}=\frac{-3-\sqrt{28}}{1} $$ = −3 − $$\sqrt{28}$$
x₁ = −3 + $$\sqrt{28}$$, x₂ = −3 − $$\sqrt{28}$$
е) 5y² + 26y − 24 = 0 5y² + 2 ⋅ 13y − 24 = 0 D = k² − ac = 13² − 5 ⋅ (−24) = 169 + 120 = 289 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-k+\sqrt{D}}{a}=\frac{-13+\sqrt{289}}{5} $$ = −13 + 17/5 = 4/5 = 0,8
x₂ = $$ \frac{-k-\sqrt{D}}{a}=\frac{-13-\sqrt{289}}{5}= $$ = −13 − 17/5 = −30/5 = −6
x₁ = 0,8, x₂ = −6
ж) z² − 34z + 289 = 0 z² − 2 ⋅ 17z + 289 = 0 D = k² − ac = (−17)² − 1 ⋅ 289 = 289 - 289 = 0 = 0, уравнение имеет 1 корень
x₁ = $$ =\frac{-k+\sqrt{D}}{a}=\frac{17+\sqrt{0}}{1} $$ = 17 + 0/1 = 17/1 = 17
x₁ = 17
з) 3x² + 32x + 80 = 0 3x² + 2 ⋅ 16x + 80 = 0 D = k² − ac = 16² − 3 ⋅ 80 = 256 - 240 = 16 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-k+\sqrt{D}}{a}=\frac{-16+\sqrt{16}}{3} $$ = −16 + 4/3 = −12/3 = −4
x₂ = $$ \frac{-k-\sqrt{D}}{a}=\frac{-16-\sqrt{16}}{3} $$ = −16 − 4/3 = −20/3 = −62/3
x₁ = −4, x₂ = −62/3