Maharashtra Board Class 10 Math Part 1 Solution Chapter 2 Practice Set 2.3 – Quadratic Equations
Balbharati Maharashtra Board Class 10 Math Part 1 Solution Chapter 2: Quadratic Equations. Marathi or English Medium Students of Class 10 get here Quadratic Equations full Practice set Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 1 Solution |
Chapter
Practice set |
Quadratic Equations
2.3 |
Practice Set 2.3
Solve the following quadratic equations by completing the square method.
(1) x2 + x – 20 = 0
Or, x2 – 4x + 5x – 20 = 0
Or, x (x – 4) + 5 (x – 4) = 0
Or, (x – 4) (x + 5) = 0
Or, (x – 4) (x + 5) = 0
Either, x – 4 = 0
Or, x = 4
Or, x + 5 = 0
Or, x = -5
Are the roots of the equation
(2) x2 + 2x – 5 = 0
Or, x2 + 2x + 1 – 6 = 0
Or, (x + 1)2 = 6
Or, x + 1 = ± √6
Or, x = ± √6 – 1
∴ Either, x = +√6 – 1
Or, x = – √6 – 1
(3) m2 – 5m = -3
Or, m2 – 5m + 3 = 0
Taking squaring method
m2 – 5m + k = (x + a)2
Or, m2 – 5m + k
= x2 + 2ax + a2
Comparing both terms we get –
2a = -5
Or, a = -5/2
∴ k = a2
Or, K = (-5/2)2
= -25/4
By equating constant – terms for the variable of m
m2 – 5m + 3 = {m + (-5/2)}2 – 25/4 + 3
= (m = 5/2)2 – 25+12/4
= (m – 5/2)2 – 13/4
Now, (m – 5/2)2 – (√13/4)2 = 0
Or, (m – 5/2 + √13/4) (m – 5/2 – √13/4)
∴ Either m – 5/2 + √15/4 = 0 Or, m – 5/2 – √13/2 = 0
Or, m = 5/2 – √13/4 = 5-√13/2 Or, m = 5/2 + √13/2
Are the m of = 5+√13/2
(4) 9y2 – 12y + 2 = 0
Or, 9y2 – 12y + 4 – 2 = 0
Or, (3y)2 – 2 × 3 × 2y + 4 – 2 = 0
Or, (3y – 2)2 – 2 = 0
Or, (3y – 2)2 = 2
Or, 3y – 2 = ± √2
Or, y = (± √2 + 2)/3
∴ Either, y = √2 + 2/3
Or, y = – √2 + 2/3
∴ Are the roots of the equation
(5) 2y2 + 9y + 10 = 0
Taking sq method it is desirable is have 1 as x2 is coefficient.
∴ Dividing with 2 on both sides
y2 + 9/2 y + 5 = 0
Now, if y2 + 9/2 y + 15 = (xy – a)2
Then, y2 + 9/2 xy + k = y2 – 2a xy + a2
Comparing the terms, y2 + 9/2 y and xy2 – 2a xy
-2a xy = 9/2 y
∴ K = a2 = (-9/4)2 = 81/96
Or, a = 9/4
Now, y2 + 9/2y + 5 – 6
Or, y2 + 9/2 y + 81/16 – 81/16 + 5 = 0 (Value of K)
Or, y2 + 2 × 9/4 × y + (9/4)2 – 81+80/16 = 0
Or, (y + 9/4)2 – 1/16 = 0
Or, (y + 9/4)2 = 1/16 or, (y + 9/4)2 = (1/4)2
∴ (y + 9/4) = ± 1/4
∴ Either, y + 9/4 = 1/9
Or, y = 1/4 – 9/4
= -8/4
= -2
Or, y + 9/4 = -1/4
Or, y = -1/4 – 9/4
= -10/4
= -5/2
Are the roots of the equation
(6) 5x2 = 4x + 7
Or, 5x2 – 4x – 7 = 0
For sq. method taking coefficient – of x as 1 is desirable
∴ Dividing 5 on both sides
x2 – 4/5x – 7/5 = 0
Now, if x2 – 4/5 x + k = (x – a)2
Then, x2 – 4/5 x + k = x2 – 2a x + a2
Comparing both terms
x2 – 4/5 x = x2 – 2ax
∴ -20x = -4/5 x a = 2/3
∴ K = a2 = (2/5)2
= 4/25
Here is your solution of Maharashtra Board Class 10 Math Part 1 Solution Chapter 2 Practice Set 2.3 – Quadratic Equations
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