Find the solution of the quadratic equation 2x 2 - mx - 25n = 0; if m + 5 = 0 and n - 1 = 0.

Given, m + 5 = 0 and n - 1 = 0 ∴ m = -5 and n = 1. Substituting values of m and n in 2x 2 - mx - 25n = 0 we get, ⇒ 2x 2 - (-5)x - 25(1) = 0 ⇒ 2x 2 + 5x - 25 = 0 ⇒ 2x 2 + 10x - 5x - 25 = 0 ⇒ 2x(x + 5) - 5(x + 5) = 0 ⇒ (2x - 5)(x + 5) = 0 ⇒ (2x - 5) = 0 or x + 5 = 0 ⇒ x = or x = -5. Hence, x = or -5.

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Show that one root of the quadratic equation x² + (3 - 2a)x - 6a = 0 is -3. Hence, find its other root.

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Solve :
(a + b)²x² - (a + b)x - 6 = 0; a + b ≠ 0.

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Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.
x² + 2(m - 1)x + (m + 5) = 0

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Mathematics

Find the solution of the quadratic equation 2x² - mx - 25n = 0; if m + 5 = 0 and n - 1 = 0.

Quadratic Equations

Related Questions

Show that one root of the quadratic equation x² + (3 - 2a)x - 6a = 0 is -3. Hence, find its other root.

Solve :
(a + b)²x² - (a + b)x - 6 = 0; a + b ≠ 0.

Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.
x² + 2(m - 1)x + (m + 5) = 0

Mathematics

Find the solution of the quadratic equation 2x2 - mx - 25n = 0; if m + 5 = 0 and n - 1 = 0.

Quadratic Equations

Related Questions

Show that one root of the quadratic equation x2 + (3 - 2a)x - 6a = 0 is -3. Hence, find its other root.

Solve :(a + b)2x2 - (a + b)x - 6 = 0; a + b ≠ 0.

Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.x2 + 2(m - 1)x + (m + 5) = 0

Find the solution of the quadratic equation 2x² - mx - 25n = 0; if m + 5 = 0 and n - 1 = 0.

Show that one root of the quadratic equation x² + (3 - 2a)x - 6a = 0 is -3. Hence, find its other root.

Solve :
(a + b)²x² - (a + b)x - 6 = 0; a + b ≠ 0.

Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.
x² + 2(m - 1)x + (m + 5) = 0