(x – 3)(2x +1) = x(x + 5) Simplifying the given equation, we get (x – 3)(2x +1) = x(x + 5) ⇒ 2x² - 6x + x - 3 = x² + 5x ⇒ x² - 10x - 3 = 0 or x² - 10x - 3 = 0 This is an equation of type ax² + bx + c = 0. Hence, the given equation is a quadratic equation.
(x – 3)(2x +1) = x(x + 5)
Simplifying the given equation, we get
(x – 3)(2x +1) = x(x + 5)
⇒ 2x² – 6x + x – 3 = x² + 5x
⇒ x² – 10x – 3 = 0
or x² – 10x – 3 = 0
This is an equation of type ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
Let, the breadth of plot = x m Therefore, the length of plot 2x + 1 m Hence, area = x(2x + 1) m² According to the questions, x(2x + 1) = 528 ⇒ 2x² + x = 528 ⇒ 2x² + x - 528 = 0 Hence, the length and breath of the plot satisfies the equation 2x² + x - 528 = 0.
Let, the breadth of plot = x m
Therefore, the length of plot 2x + 1 m
Hence, area = x(2x + 1) m²
According to the questions, x(2x + 1) = 528
⇒ 2x² + x = 528
⇒ 2x² + x – 528 = 0
Hence, the length and breath of the plot satisfies the equation 2x² + x – 528 = 0.
(x + 2)³ = 2x (x² – 1) Simplifying the given equation, we get (x + 2)³ = 2x (x² – 1) ⇒ x³ + 6x² + 12x + 8 = 2x³ - 2x ⇒ - x³ + 6x² + 14x + 8 = 0 or x³ - 6x² - 14x - 8 = 0 This is not an equation of type ax² + bx + c = 0. Hence, the given equation is a quadratic equation.
(x + 2)³ = 2x (x² – 1)
Simplifying the given equation, we get
(x + 2)³ = 2x (x² – 1)
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x
⇒ – x³ + 6x² + 14x + 8 = 0
or x³ – 6x² – 14x – 8 = 0
This is not an equation of type ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
Let the first integer = x therefore, the second integer = x + 1 Hence, the product = x(x + 1) According to the questions, x(x + 1) = 306 ⇒ x² + x = 306 ⇒ x² + x - 306 = 0 Hence, the two consecutive integers satisfies quadratic equation x² + x - 306 = 0
Let the first integer = x
therefore, the second integer = x + 1
Hence, the product = x(x + 1)
According to the questions, x(x + 1) = 306
⇒ x² + x = 306
⇒ x² + x – 306 = 0
Hence, the two consecutive integers satisfies quadratic equation x² + x – 306 = 0
Check whether the following is quadratic equation: (x – 3)(2x +1) = x(x + 5)
(x – 3)(2x +1) = x(x + 5) Simplifying the given equation, we get (x – 3)(2x +1) = x(x + 5) ⇒ 2x² - 6x + x - 3 = x² + 5x ⇒ x² - 10x - 3 = 0 or x² - 10x - 3 = 0 This is an equation of type ax² + bx + c = 0. Hence, the given equation is a quadratic equation.
(x – 3)(2x +1) = x(x + 5)
Simplifying the given equation, we get
(x – 3)(2x +1) = x(x + 5)
⇒ 2x² – 6x + x – 3 = x² + 5x
⇒ x² – 10x – 3 = 0
or x² – 10x – 3 = 0
This is an equation of type ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
See lessRepresent the situations in the form of quadratic equations : The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Let, the breadth of plot = x m Therefore, the length of plot 2x + 1 m Hence, area = x(2x + 1) m² According to the questions, x(2x + 1) = 528 ⇒ 2x² + x = 528 ⇒ 2x² + x - 528 = 0 Hence, the length and breath of the plot satisfies the equation 2x² + x - 528 = 0.
Let, the breadth of plot = x m
Therefore, the length of plot 2x + 1 m
Hence, area = x(2x + 1) m²
According to the questions, x(2x + 1) = 528
⇒ 2x² + x = 528
⇒ 2x² + x – 528 = 0
Hence, the length and breath of the plot satisfies the equation 2x² + x – 528 = 0.
See lessCheck whether the following is quadratic equations : (x + 2)³ = 2x (x² – 1)
(x + 2)³ = 2x (x² – 1) Simplifying the given equation, we get (x + 2)³ = 2x (x² – 1) ⇒ x³ + 6x² + 12x + 8 = 2x³ - 2x ⇒ - x³ + 6x² + 14x + 8 = 0 or x³ - 6x² - 14x - 8 = 0 This is not an equation of type ax² + bx + c = 0. Hence, the given equation is a quadratic equation.
(x + 2)³ = 2x (x² – 1)
See lessSimplifying the given equation, we get
(x + 2)³ = 2x (x² – 1)
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x
⇒ – x³ + 6x² + 14x + 8 = 0
or x³ – 6x² – 14x – 8 = 0
This is not an equation of type ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
Represent the situations in the form of quadratic equations : The product of two consecutive positive integers is 306. We need to find the integers.
Let the first integer = x therefore, the second integer = x + 1 Hence, the product = x(x + 1) According to the questions, x(x + 1) = 306 ⇒ x² + x = 306 ⇒ x² + x - 306 = 0 Hence, the two consecutive integers satisfies quadratic equation x² + x - 306 = 0
Let the first integer = x
See lesstherefore, the second integer = x + 1
Hence, the product = x(x + 1)
According to the questions, x(x + 1) = 306
⇒ x² + x = 306
⇒ x² + x – 306 = 0
Hence, the two consecutive integers satisfies quadratic equation x² + x – 306 = 0
Find the roots of the following quadratic equations by factorisation: 2x² + x – 6 = 0
2x² + x – 6 = 0 Solving the quadratic equation, we get 2x² + x – 6 = 0 ⇒ 2x² - 4x + 3x - 6 = 0 ⇒ 2x(x - 2) +3 (x - 2) = 0 ⇒ (x - 2) (2x + 3) = 0 ⇒ (x - 2) = 0 or (2x + 3) = 0 Either x = 2 or x = - 3/2 Hence, the roots of the given quadratic equation are 2 and - 3/2.
2x² + x – 6 = 0
See lessSolving the quadratic equation, we get
2x² + x – 6 = 0 ⇒ 2x² – 4x + 3x – 6 = 0
⇒ 2x(x – 2) +3 (x – 2) = 0
⇒ (x – 2) (2x + 3) = 0
⇒ (x – 2) = 0 or (2x + 3) = 0
Either x = 2 or x = – 3/2
Hence, the roots of the given quadratic equation are 2 and – 3/2.