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Introduction to Quadratic Equations Video Lecture | Mathematics (Maths) Class 10
FAQs on Introduction to Quadratic Equations Video Lecture - Mathematics (Maths) Class 10
| 1. What is a quadratic equation and how is it generally represented? |  |
Ans. A quadratic equation is a polynomial equation of the second degree, typically represented in the standard form as ax² + bx + c = 0, where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. The solutions to quadratic equations can be found using various methods including factoring, completing the square, and the quadratic formula.
| 2. What are the different methods to solve quadratic equations? | |
Ans. There are several methods to solve quadratic equations:
1. <b>Factoring</b>: If the quadratic can be factored into two binomials, the solutions can be found by setting each factor to zero.
2. <b>Completing the Square</b>: This method involves rearranging the equation into a perfect square trinomial.
3. <b>Quadratic Formula</b>: The most universally applicable method is using the formula x = (-b ± √(b² - 4ac)) / (2a), which gives the roots directly.
4. <b>Graphing</b>: Quadratic equations can also be solved graphically by identifying the points where the parabola intersects the x-axis.
| 3. What is the significance of the discriminant in a quadratic equation? | |
Ans. The discriminant, represented as D = b² - 4ac, is a key component of a quadratic equation. It helps determine the nature of the roots:
- If D > 0, there are two distinct real roots.
- If D = 0, there is exactly one real root (a repeated root).
- If D < 0, there are no real roots, but two complex roots. This information is crucial for understanding the solutions of the equation.
| 4. Can all quadratic equations be solved by the quadratic formula? | |
Ans. Yes, all quadratic equations can be solved using the quadratic formula, regardless of whether they can be factored or not. The formula can yield real or complex solutions based on the value of the discriminant. This universality makes it a reliable method for solving any quadratic equation.
| 5. What are some real-world applications of quadratic equations? | |
Ans. Quadratic equations have numerous real-world applications, including:
1. <b>Physics</b>: They are used to model projectile motion and the path of objects under the influence of gravity.
2. <b>Finance</b>: Quadratic equations can help in calculating profit maximization and cost minimization scenarios.
3. <b>Engineering</b>: They appear in calculations involving areas, structural designs, and optimization problems.
4. <b>Biology</b>: Quadratic models can describe population growth under certain constraints.
These applications illustrate the importance of quadratic equations across various fields.
Text Transcript from Video
0:0:0: did you know the word quadratic was
0:0:3: derived from a Latin word which means
0:0:5: Square so in algebra we use quadratic
0:0:9: equation to refer to any equation with
0:0:11: the highest power of two for
0:0:13: examples 2x² + 3 x 1 = 0 x² + 7 = 0 5 x²
0:0:24: equal to
0:0:25: 0 imagine you have a rectangle with area
0:0:29: equal to 300 M squ and sides equal to 2x
0:0:33: + 1 and X let's try to make an
0:0:36: expression out of it we all know the
0:0:39: formula to calculate the area of a
0:0:41: rectangle which is length multiplied by
0:0:44: breadth so here we get 2 x + 1 * X on
0:0:50: multiplying we get 2 x² + x also because
0:0:55: we know that the area is equal to 300 M
0:0:59: Square we can write WR 2x² + x = 300
0:1:4: m² what do you think is this a quadratic
0:1:8: equation let's
0:1:10: analyze first is this an
0:1:13: equation yes it is why because it has an
0:1:18: equal sign in it now what do you think
0:1:22: is it a quadratic
0:1:25: equation yes you are absolutely right it
0:1:29: is a quadratic
0:1:30: equation why as we studied a quadratic
0:1:34: equation has the highest power of a
0:1:36: variable equal to two now let's
0:1:40: generalize the quadratic equation if we
0:1:43: put a instead of 2 b instead of 1 and C
0:1:46: instead of 300 we get a x² + b x + C is
0:1:50: equal to 0 this is called the standard
0:1:54: form of a quadratic equation with a
0:1:57: condition that a cannot be equal to zero
0:2:1: why because if a becomes equal to zero
0:2:4: the term of x² will also disappear and
0:2:7: the basic definition of quadratic
0:2:9: equation that the equation has highest
0:2:11: power of to will not be
0:2:13: satisfied so always remember a quadratic
0:2:18: equation is always of the form a x² + b
0:2:21: x + C is equal to 0 where a is not equal
0:2:24: to zero let's now try some questions
0:2:27: from exercise
0:2:28: 4.1 the first question says check
0:2:32: whether the following are quadratic
0:2:35: equations x + 1 s = to 2 * x - 3 do you
0:2:42: think this is a quadratic equation does
0:2:45: it satisfy the condition let's try
0:2:49: together now here we need to always be
0:2:53: careful before judging anything the
0:2:55: first task should always be simplifying
0:2:58: the equation here we need to open up the
0:3:2: brackets x + 1 s isn't it similar to an
0:3:7: identity we have
0:3:8: studied which one is it yes it is A + B
0:3:13: squ which is equal to a² + b² + 2 a
0:3:19: similarly we open this bracket we get x²
0:3:23: + 1 s + 2 * x * 1 and on the right hand
0:3:28: side we have 2 * x - 3 opening this
0:3:33: bracket we get 2 x - 2 * 3 that is 6 so
0:3:39: equating LHS and rhs we get x² + 1 + 2x
0:3:44: = 2 x - 6 on solving this equation we
0:3:49: get x² + 7 = 0 here we can see that the
0:3:54: highest power of the variable X is 2 so
0:3:59: it is a quad quadratic
0:4:0: equation so while solving these types of
0:4:3: questions always remember to First
0:4:6: simplify the expression and then judge
0:4:8: whether it is a quadratic equation or
0:4:10: not let's move on to the next question
0:4:14: that is represent the following
0:4:16: situations in the form of quadratic
0:4:19: equations here's a tip to tackle these
0:4:21: type of questions always write down the
0:4:24: information as you read the
0:4:26: question let's start the a area of a
0:4:30: rectangular plot is 528 m squ so we note
0:4:35: down that the area of the rectangle is
0:4:37: equal to 528 m squar the length of the
0:4:42: plot in met is one more than twice its
0:4:45: breadth here if we take breadth to be X
0:4:49: therefore the length comes out to be 2x
0:4:52: + 1 we need to find the length and
0:4:55: breadth of the
0:4:56: plot so we have length equal 2x + 1 and
0:5:1: bread equal to X and area equal to 528
0:5:6: m² also we know the formula for area of
0:5:9: rectangle that is length into breadth =
0:5:12: 2x + 1 into x = 528
0:5:17: m² therefore the equation comes out to
0:5:20: be 2x² + x - 528 is equal to 0 and this
0:5:25: is indeed a quadratic equation do try
0:5:29: the rest of the questions on your
0:5:31: own now that we have understood how to
0:5:34: form a quadratic equation we will move
0:5:36: on to the next section which is solution
0:5:38: of a quadratic equation by
0:5:41: factorization consider a quadratic
0:5:43: equation 2x^2 - 3x + 1 = 0 if we replace
0:5:49: X by 1 on the left hand side we get 2
0:5:53: into 1 2 - 3 into 1 + 1 which turns out
0:5:56: to be 0er which is the right hand side
0:6:0: this implies one is a root of the
0:6:2: quadratic
0:6:3: equation so in general a real number
0:6:6: Alpha is called the root of a quadratic
0:6:8: equation only if it satisfies the
0:6:11: equation the method of factorization we
0:6:14: are going to study is called middle term
0:6:17: splitting let us consider the standard
0:6:19: form of quadratic equation that is ax² +
0:6:23: BX + c equal 0 in this method we have to
0:6:28: split the middle number which is B into
0:6:30: two numbers p and Q such that p + Q is
0:6:34: equal to B and P into Q is equal to a
0:6:37: multiplied by C let's try a question to
0:6:41: better understand the
0:6:42: concept find the roots of the quadratic
0:6:45: equation 6 x^2 - x - 2 = 0 so here if we
0:6:52: compare this quadratic equation with the
0:6:54: standard form a x² + b x + C = 0 we get
0:6:59: get = to 6 B = -1 and C = to
0:7:4: -2 so for splitting middle term method
0:7:7: we need to split minus1 in two numbers p
0:7:10: and Q such that p + Q = to min-1 and P
0:7:14: into Q = to a into C = 6 into -2 =
0:7:20: -12 think yourself P can be minus 4 and
0:7:24: Q can be 3 let's check once if these
0:7:28: values satisfy the
0:7:29: conditions p + Q = - 4 + 3 = to -1 and
0:7:35: it is indeed equal to B P into Q = -4
0:7:40: into 3 = -12 and this is equal to a into
0:7:44: C = to
0:7:46: -2 so we write 6 x 2 - 4x + 3x - 2 = 0
0:7:54: next step is to take out common
0:7:56: numbers so we take out 2 x X common from
0:8:0: the first two terms and one from the
0:8:2: last two terms we get 2x into 3x - 2 + 1
0:8:7: into 3x - 2 again we see 3x - 2 to be
0:8:13: common we take that also out and get 3x
0:8:17: - 2 * 2x + 1 so we get two values of X
0:8:23: by solving 3x - 2 and 2 x + 1 which are
0:8:27: -1 by 2 and 2 by 3 hence the roots of
0:8:32: the quadratic equation are -1x 2 and 2x
0:8:36: 3 we hope that you have understood the
0:8:38: concept of quadratic
0:8:40: equations now here's a secret on how you
0:8:43: can study effectively with the urv app
0:8:47: you can learn with chapter notes watch
0:8:49: video lectures and solve ncer based McQ
0:8:53: tests of this chapter on
0:8:55: UV and that's not all you also get
0:8:58: amazing courses for class 10 and much
0:9:0: more for your class 10
0:9:2: preparation thank you
0:0:3: derived from a Latin word which means
0:0:5: Square so in algebra we use quadratic
0:0:9: equation to refer to any equation with
0:0:11: the highest power of two for
0:0:13: examples 2x² + 3 x 1 = 0 x² + 7 = 0 5 x²
0:0:24: equal to
0:0:25: 0 imagine you have a rectangle with area
0:0:29: equal to 300 M squ and sides equal to 2x
0:0:33: + 1 and X let's try to make an
0:0:36: expression out of it we all know the
0:0:39: formula to calculate the area of a
0:0:41: rectangle which is length multiplied by
0:0:44: breadth so here we get 2 x + 1 * X on
0:0:50: multiplying we get 2 x² + x also because
0:0:55: we know that the area is equal to 300 M
0:0:59: Square we can write WR 2x² + x = 300
0:1:4: m² what do you think is this a quadratic
0:1:8: equation let's
0:1:10: analyze first is this an
0:1:13: equation yes it is why because it has an
0:1:18: equal sign in it now what do you think
0:1:22: is it a quadratic
0:1:25: equation yes you are absolutely right it
0:1:29: is a quadratic
0:1:30: equation why as we studied a quadratic
0:1:34: equation has the highest power of a
0:1:36: variable equal to two now let's
0:1:40: generalize the quadratic equation if we
0:1:43: put a instead of 2 b instead of 1 and C
0:1:46: instead of 300 we get a x² + b x + C is
0:1:50: equal to 0 this is called the standard
0:1:54: form of a quadratic equation with a
0:1:57: condition that a cannot be equal to zero
0:2:1: why because if a becomes equal to zero
0:2:4: the term of x² will also disappear and
0:2:7: the basic definition of quadratic
0:2:9: equation that the equation has highest
0:2:11: power of to will not be
0:2:13: satisfied so always remember a quadratic
0:2:18: equation is always of the form a x² + b
0:2:21: x + C is equal to 0 where a is not equal
0:2:24: to zero let's now try some questions
0:2:27: from exercise
0:2:28: 4.1 the first question says check
0:2:32: whether the following are quadratic
0:2:35: equations x + 1 s = to 2 * x - 3 do you
0:2:42: think this is a quadratic equation does
0:2:45: it satisfy the condition let's try
0:2:49: together now here we need to always be
0:2:53: careful before judging anything the
0:2:55: first task should always be simplifying
0:2:58: the equation here we need to open up the
0:3:2: brackets x + 1 s isn't it similar to an
0:3:7: identity we have
0:3:8: studied which one is it yes it is A + B
0:3:13: squ which is equal to a² + b² + 2 a
0:3:19: similarly we open this bracket we get x²
0:3:23: + 1 s + 2 * x * 1 and on the right hand
0:3:28: side we have 2 * x - 3 opening this
0:3:33: bracket we get 2 x - 2 * 3 that is 6 so
0:3:39: equating LHS and rhs we get x² + 1 + 2x
0:3:44: = 2 x - 6 on solving this equation we
0:3:49: get x² + 7 = 0 here we can see that the
0:3:54: highest power of the variable X is 2 so
0:3:59: it is a quad quadratic
0:4:0: equation so while solving these types of
0:4:3: questions always remember to First
0:4:6: simplify the expression and then judge
0:4:8: whether it is a quadratic equation or
0:4:10: not let's move on to the next question
0:4:14: that is represent the following
0:4:16: situations in the form of quadratic
0:4:19: equations here's a tip to tackle these
0:4:21: type of questions always write down the
0:4:24: information as you read the
0:4:26: question let's start the a area of a
0:4:30: rectangular plot is 528 m squ so we note
0:4:35: down that the area of the rectangle is
0:4:37: equal to 528 m squar the length of the
0:4:42: plot in met is one more than twice its
0:4:45: breadth here if we take breadth to be X
0:4:49: therefore the length comes out to be 2x
0:4:52: + 1 we need to find the length and
0:4:55: breadth of the
0:4:56: plot so we have length equal 2x + 1 and
0:5:1: bread equal to X and area equal to 528
0:5:6: m² also we know the formula for area of
0:5:9: rectangle that is length into breadth =
0:5:12: 2x + 1 into x = 528
0:5:17: m² therefore the equation comes out to
0:5:20: be 2x² + x - 528 is equal to 0 and this
0:5:25: is indeed a quadratic equation do try
0:5:29: the rest of the questions on your
0:5:31: own now that we have understood how to
0:5:34: form a quadratic equation we will move
0:5:36: on to the next section which is solution
0:5:38: of a quadratic equation by
0:5:41: factorization consider a quadratic
0:5:43: equation 2x^2 - 3x + 1 = 0 if we replace
0:5:49: X by 1 on the left hand side we get 2
0:5:53: into 1 2 - 3 into 1 + 1 which turns out
0:5:56: to be 0er which is the right hand side
0:6:0: this implies one is a root of the
0:6:2: quadratic
0:6:3: equation so in general a real number
0:6:6: Alpha is called the root of a quadratic
0:6:8: equation only if it satisfies the
0:6:11: equation the method of factorization we
0:6:14: are going to study is called middle term
0:6:17: splitting let us consider the standard
0:6:19: form of quadratic equation that is ax² +
0:6:23: BX + c equal 0 in this method we have to
0:6:28: split the middle number which is B into
0:6:30: two numbers p and Q such that p + Q is
0:6:34: equal to B and P into Q is equal to a
0:6:37: multiplied by C let's try a question to
0:6:41: better understand the
0:6:42: concept find the roots of the quadratic
0:6:45: equation 6 x^2 - x - 2 = 0 so here if we
0:6:52: compare this quadratic equation with the
0:6:54: standard form a x² + b x + C = 0 we get
0:6:59: get = to 6 B = -1 and C = to
0:7:4: -2 so for splitting middle term method
0:7:7: we need to split minus1 in two numbers p
0:7:10: and Q such that p + Q = to min-1 and P
0:7:14: into Q = to a into C = 6 into -2 =
0:7:20: -12 think yourself P can be minus 4 and
0:7:24: Q can be 3 let's check once if these
0:7:28: values satisfy the
0:7:29: conditions p + Q = - 4 + 3 = to -1 and
0:7:35: it is indeed equal to B P into Q = -4
0:7:40: into 3 = -12 and this is equal to a into
0:7:44: C = to
0:7:46: -2 so we write 6 x 2 - 4x + 3x - 2 = 0
0:7:54: next step is to take out common
0:7:56: numbers so we take out 2 x X common from
0:8:0: the first two terms and one from the
0:8:2: last two terms we get 2x into 3x - 2 + 1
0:8:7: into 3x - 2 again we see 3x - 2 to be
0:8:13: common we take that also out and get 3x
0:8:17: - 2 * 2x + 1 so we get two values of X
0:8:23: by solving 3x - 2 and 2 x + 1 which are
0:8:27: -1 by 2 and 2 by 3 hence the roots of
0:8:32: the quadratic equation are -1x 2 and 2x
0:8:36: 3 we hope that you have understood the
0:8:38: concept of quadratic
0:8:40: equations now here's a secret on how you
0:8:43: can study effectively with the urv app
0:8:47: you can learn with chapter notes watch
0:8:49: video lectures and solve ncer based McQ
0:8:53: tests of this chapter on
0:8:55: UV and that's not all you also get
0:8:58: amazing courses for class 10 and much
0:9:0: more for your class 10
0:9:2: preparation thank you
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