The solution set for the equation, 25x(x + 1) = -4, is: 1. {1/5, 4/5} 2. {-1/5, 4/5} 3. {-4/5, -1/5} 4. {-4/5, 1/5}

⇒ 25x(x + 1) = -4 ⇒ 25x2 + 25x + 4 = 0 ⇒ 25x2 + 5x + 20x + 4 = 0 ⇒ 5x(5x + 1) + 4(5x + 1) = 0 ⇒ (5x + 4)(5x + 1) = 0 ⇒ (5x + 4) = 0 or (5x + 1) = 0 [Using Zero-product rule] ⇒ 5x = -4 or 5x = -1 ⇒ x = or x = . The solution set: . Hence, option 3 is the correct option.

Mathematics

The solution set for the equation, 25x(x + 1) = -4, is:
$\Big{\dfrac{1}{5}, \dfrac{4}{5}\Big}$
$\Big{-\dfrac{1}{5}, \dfrac{4}{5}\Big}$
$\Big{-\dfrac{4}{5}, -\dfrac{1}{5}\Big}$
$\Big{-\dfrac{4}{5}, \dfrac{1}{5}\Big}$

Quadratic Equations

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Answer

⇒ 25x(x + 1) = -4

⇒ 25x² + 25x + 4 = 0

⇒ 25x² + 5x + 20x + 4 = 0

⇒ 5x(5x + 1) + 4(5x + 1) = 0

⇒ (5x + 4)(5x + 1) = 0

⇒ (5x + 4) = 0 or (5x + 1) = 0 [Using Zero-product rule]

⇒ 5x = -4 or 5x = -1

⇒ x = $\dfrac{-4}{5}$ or x = $\dfrac{-1}{5}$ .

The solution set: $\Big{-\dfrac{4}{5}, -\dfrac{1}{5}\Big}$ .

Hence, option 3 is the correct option.

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Mathematics

The solution set for the equation, 25x(x + 1) = -4, is:
$\Big{\dfrac{1}{5}, \dfrac{4}{5}\Big}$
$\Big{-\dfrac{1}{5}, \dfrac{4}{5}\Big}$
$\Big{-\dfrac{4}{5}, -\dfrac{1}{5}\Big}$
$\Big{-\dfrac{4}{5}, \dfrac{1}{5}\Big}$

Quadratic Equations

Answer

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Mathematics

Quadratic Equations

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The value of the discriminant of the equation, 3x² = - 11x - 10, is:
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If -5 is a root of the quadratic equation 2x² + px - 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, then the value of k is:
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