NCERT Class 10th Mathematics
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NCERT Class 10th – Chapter 4 – Exercise 4.1 Quadratic Equations

NCERT full solutions  here you can find full detailed solutions for all the questions  with proper format. class 10 maths ncert solutions chapter 4 exercise 4.1.

You can see the solution for complete chapter here –

EXERCISE:-4.1

QUE:-1. 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

SOL:-

(i)(x + 1)²= 2(x – 3)

by solving the equation we get

x² +2x +1 =2x -6

x² +2x +1 -2x +6 =0

x² +0x +7=0

the equation is in the form ax² +bx +c =0

∴ equation is a quadratic equations

(ii) x²– 2x = (–2) (3 – x)

by solving the equation we get

x²– 2x = (–2) (3 – x)

x²– 2x = -6 +2x

x²+0x +6 =0

the equation is in the form ax² +bx +c =0

∴ equation is a quadratic equations

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

by solving the equation we get

x² +x -2x -2 =x² +3x -x -3

x² +x -2x -2 -x²-3x +x +3=0

-3x +1 =0

the equation is in not the form ax² +bx +c =0

∴ equation is not a quadratic equations

(iv) (x – 3)(2x +1) = x(x + 5)

by solving the equation we get

(x – 3)(2x +1) = x(x + 5)

2x² +x -6x -3 =x² +5x

2x² +x -6x -3-x² -5x=0

x² -10x-3=0

the equation is in the form ax² +bx +c =0

∴ equation is a quadratic equations

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

by solving the equation we get

2x² -6x -x +3 = x² -x +5x -5

x² – 11x +8 =0

the equation is in the form ax² +bx +c =0

∴ equation is a quadratic equations

(vi) x²+ 3x + 1 = (x – 2)²

by solving the equation we get

x²+ 3x + 1 = (x – 2)²

x²+ 3x + 1 = x² -4x+4

7x -3 =0

the equation is in not the form ax² +bx +c =0

∴ equation is not a quadratic equations

(vii) (x + 2)³= 2x (x²– 1)

by solving the equation we get

(x + 2)³= 2x (x²– 1)

x³ +6x² +12x +8 =2x³ -2x

-x³ +6x² +14x +8 =0

the equation is in not the form ax² +bx +c =0

∴ equation is not a quadratic equations

(viii) x³– 4x²– x + 1 = (x – 2)³

by solving the equation we get

x³– 4x²– x + 1 = (x – 2)³

x³– 4x²– x + 1 = x³ -6x² +12x-8

2x² -13x +9 =0

the equation is in the form ax² +bx +c =0

∴ equation is a quadratic equations

QUE:-2. Represent the following situations in the form of quadratic equations :
     (i) 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.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years)
    3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been
    8 km/h less, then it would have taken 3 hours more to cover the same distance. We
     need to find the speed of the train.

SOL:-

(i) let the beadth of the plot is x m

and the length of plot 2x +1(as state in que.)

area of plot (L ×B) = x (2x +1)

2x² +x =528

2x² +x -528=0

so length and breadth of the plot is satisfies the quadratic equ. 2x² +x -528=0

(ii) let the first number is x

and the second number is x+1

according to the question

x(x+1) =306

x² +x =306

x²+x -306 =0

so length and breadth of the plot is satisfies the quadratic equ. x² +x -306=0

(iii)  let the age of rohan x

age of his mother x +26

after 3 year age of Rohan x +3 and his mother x +29

product of there age (x+3)(x+29)=360

x² +3x +29x +87=360

x² +32x -273=0

so age of Rohan and his mother’s age is satisfies the quadratic equ. x² +32x -273=0

(iv)let the speed of train x

if speed less 8 km/h x-8

480/x-8-480/x=3

480x -480x +3640=3(x-8)x

3x² -24x -3640=0

so the speed of train is satisfies the quadratic equ. 3x² -24x -3640=0

You can see the solution for complete chapter here –

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