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Solusi - Memecahkan persamaan kuadrat dengan memfaktorkan

Bentuk tepat:

x = - \frac{7}{2}

x=-\frac{7}{2}

Bentuk desimal:

x = - 3, 5

x=-3,5

Persamaan dalam bentuk faktor:

{(2 x + 7)}^{2} = 0

(2x+7)^2=0

Lihat langkah

Penjelasan langkah demi langkah

1. Pastikan bahwa persamaan adalah trinomial kuadrat sempurna

In a perfect square trinomial, the rule is that the square root of coefficient $a$ times the square root of coefficient $c$ times two equals coefficient $b$ :
$\sqrt{a} \cdot \sqrt{c} \cdot 2 = b$

Para encontrar los coeficientes, usa la forma estándar de una ecuación cuadrática:
$a x^{2} + b x + c = 0$
$4 x^{2} + 28 x + 49 = 0$

Coeficiente $a$ $= 4$
Coeficiente $b$ $= 28$
Coeficiente $c$ $= 49$

Introduce los coeficientes en la regla y verifica si es cierto:

$\sqrt{(4)} \cdot \sqrt{(49)} \cdot 2 = 28$

Ambil akar kuadrat

$2 \cdot 7 \cdot 2 = 28$

Sederhanakan ekspresi

$28 = 28$

Because the equation is true,
$4 x^{2} + 28 x + 49$
is a perfect square trinomial.

2. Temukan faktor dari trinomial kuadrat sempurna

Untuk mencari faktor dari trinomial kuadrat sempurna:
$4 x^{2} + 28 x + 49$
Gunakan rumus trinomial kuadrat sempurna:
$a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$

$a^{2} = 4 x^{2}$

$b^{2} = 49$

Ambil akar kuadrat

$a = \sqrt{4 x^{2}}$

$b = \sqrt{49}$

Sederhanakan ekspresi

$a = 2 x$

$b = 7$

${(a + b)}^{2} = {(2 x + 7)}^{2}$
Faktor dari $4 x^{2} + 28 x + 49$ adalah $(2 x + 7)$

3. Temukan akar dari persamaan kuadrat

Cari akar dari:
$4 x^{2} + 28 x + 49 = 0$
dengan menggunakan bentuk faktornya:
${(2 x + 7)}^{2} = 0$

Si
${(2 x + 7)}^{2} = 0$
Entonces
$(2 x + 7) (2 x + 7) = 0$
Lo que significa
$(2 x + 7) = 0$

Resuelve para $x$ :

4 tambahan langkah

$(2 x + 7) = 0$

Kurangi dari kedua ruas:

$(2 x + 7) - 7 = 0 - 7$

Sederhanakan hitungan:

$2 x = 0 - 7$

Sederhanakan hitungan:

$2 x = - 7$

Bagi kedua ruas dengan :

$\frac{(2 x)}{2} = \frac{- 7}{2}$

Sederhanakan pecahan:

$x = \frac{- 7}{2}$

4. Graf

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