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Equation With Question For Practise
| 1. | Board | CBSE |
| 2. | Textbook | NCERT |
| 3. | Class | Class 10 |
| 4. | Subject | Notes |
| 5. | Chapter | Chapter 4 |
| 6. | Chapter Name | Quadratic Equations |
| 7. | Category | CBSE Revision Notes |
Check whether the following are quadratic equations:
(i) (x + 1)2 = 2(x – 3)
(ii) x2 – 2x = (–2) (3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
(iv) (x – 3)(2x +1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
(vi) x2 + 3x + 1 = (x – 2)2
(vii) (x + 2)3 = 2x (x2 – 1)
(viii) x3 – 4x2 – x + 1 = (x – 2)3
Solutions:
(i) Given,
(x + 1)2 = 2(x – 3)
By using the formula for (a+b)2 = a2+2ab+b2
⇒ x2 + 2x + 1 = 2x – 6
⇒ x2 + 7 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(ii) Given, x2 – 2x = (–2) (3 – x)
⇒ x2 – 2x = -6 + 2x
⇒ x2 – 4x + 6 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3)
By multiplication
⇒ x2 – x – 2 = x2 + 2x – 3
⇒ 3x – 1 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
(iv) Given, (x – 3)(2x +1) = x(x + 5)
By multiplication
⇒ 2x2 – 5x – 3 = x2 + 5x
⇒ x2 – 10x – 3 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1)
By multiplication
⇒ 2x2 – 7x + 3 = x2 + 4x – 5
⇒ x2 – 11x + 8 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(vi) Given, x2 + 3x + 1 = (x – 2)2
By using the formula for (a-b)2=a2-2ab+b2
⇒ x2 + 3x + 1 = x2 + 4 – 4x
⇒ 7x – 3 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
(vii) Given, (x + 2)3 = 2x(x2 – 1)
By using the formula for (a+b)3 = a3+b3+3ab(a+b)
⇒ x3 + 8 + x2 + 12x = 2x3 – 2x
⇒ x3 + 14x – 6x2 – 8 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(viii) Given, x3 – 4x2 – x + 1 = (x – 2)3
By using the formula for (a-b)3 = a3-b3-3ab(a-b)
⇒ x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x
⇒ 2x2 – 13x + 9 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Solution of a Quadratic Equation by Factorisation
Consider the quadratic equation 2x²2 3x + 1 = 0. If we replace x by 1 on the LHS of this equation, we get (2 x 1²) – (3 x 1) + 1 = 0 = RHS the equation.
We say that I am a root of the quadratic equation 2x²-3x+1=0. This also means that 1 is a zero of the quadratic polynomial 2x² – 3x + 1.
In general, a real number a is called a root of the quadratic equation ar²+bx+c= 0, a 0 if a a² + ba+c= 0. We also say that x = a is a solution of the quadratic equation, or that a satisfies the quadratic equation.
Note that the zeroes of the quadratic polynomial ar²+bx+c and the roots of the quadratic equation ar+bx+c = 0 are the same.
You have observed, in Chapter 2, that a quadratic polynomial can have at most two zeroes. So, any quadratic equation can have almost two roots.
You have learned in Class IX, how to factorize quadratic polynomials by splitting their middle terms. We shall use this knowledge for finding the roots of a quadratic equation. Let us see how.
Example 3: Find the roots of the equation 2²-5x +3=0, by factorisation.
Solution : Let us first split the middle term – 5.ras-2r-3x [because (-21) × (-3x) =
6 = (2x²) x 3]. So, 2²-5x + 3 = 2²-2x − 3x + 3 = 2x (x – 1)-3(x-1)= (2x – 3)(x – 1)
Now, 2r-5x +3=0 can be rewritten as (2x-3)(x-1)=0. So, the values of x for which 2-5x+3=0 are the same for which (2x-3)(x-1)=0, ie., either 2x-3-0 orx-1=0.
Now, 2x-3=0 gives = and x-1=0 gives.x = 1. 3 So, X and x = 1 are the solutions of the equation. In other words, I and I are the roots of the equation 2r-5x +3=0. Verify that these are the roots of the given equation.
| Author | – |
| Language | English |
| No. of Pages | 23 |
| PDF Size | 1 MB |
| Category | Subject |
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