(x+1)² = 2(x - 3) simplifying the given equation, we got (x + 1)² = 2(x-3) ⇒ x² + 2x + 1 = 2x - 6 ⇒ x ² + 7 = 0 or x² +0x + 7 = 0 this is an equation of type ax² + bx + c = 0 Hence, the given equation is a quadratic equation. video explanation
(x+1)² = 2(x – 3)
simplifying the given equation, we got
(x + 1)² = 2(x-3)
⇒ x² + 2x + 1 = 2x – 6
⇒ x ² + 7 = 0
or x² +0x + 7 = 0
this is an equation of type ax² + bx + c = 0
Hence, the given equation is a quadratic equation.
Find the roots of the following quadratic equations by factorisation: 2x² – x +1/8 = 0
2x² – x + 1/8 = 0 Solving the quadratic equation, we got 16x² - 8x + 1 = 0 ⇒ 16x² - 4x - 4x + 1 = 0 ⇒ 4x(4x -1) - 1(4x -1) = 0 ⇒ (4x - 1) (4x -1) = 0 ⇒ (4x - 1) = 0 or (4x - 1) = 0 Either x = 1/4 or x = 1/4 Hence, the roots of given quadratic equation are 1/4 and 1/4.
2x² – x + 1/8 = 0
See lessSolving the quadratic equation, we got
16x² – 8x + 1 = 0
⇒ 16x² – 4x – 4x + 1 = 0
⇒ 4x(4x -1) – 1(4x -1) = 0
⇒ (4x – 1) (4x -1) = 0
⇒ (4x – 1) = 0 or (4x – 1) = 0
Either x = 1/4 or x = 1/4
Hence, the roots of given quadratic equation are 1/4 and 1/4.
Check whether the following are quadratic equations : (x + 1)² = 2(x – 3)
(x+1)² = 2(x - 3) simplifying the given equation, we got (x + 1)² = 2(x-3) ⇒ x² + 2x + 1 = 2x - 6 ⇒ x ² + 7 = 0 or x² +0x + 7 = 0 this is an equation of type ax² + bx + c = 0 Hence, the given equation is a quadratic equation. video explanation
(x+1)² = 2(x – 3)
simplifying the given equation, we got
(x + 1)² = 2(x-3)
⇒ x² + 2x + 1 = 2x – 6
⇒ x ² + 7 = 0
or x² +0x + 7 = 0
this is an equation of type ax² + bx + c = 0
Hence, the given equation is a quadratic equation.
video explanation
See less