Quadratic Equations-Exercise 4.1-Solution-Class 10-Mathematics

byAmlesh Mahato •October 11, 2020

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Exercise:4.1

1. Check whether the following are quadratic equations :

(i) $\large (x + 1)^2=2(x-3)$

Solution:

$\large (x + 1)^2=2(x-3)$

$\large \Rightarrow x^{2} + 1^2+2x=2x-6$

$\large \Rightarrow x^{2} + 1^+2x-2x+6=0$

$\large \Rightarrow x^{2}+1+6=0$

⇨ $\large x^2$ + 7 = 0

Yes, It match the Quadratic Equation Format.

∵ $\large ax^{2}+bx+c=0$

(ii) $\large x^2-2x=(-2)(3-x)$

Solution:

$\large x^2-2x=(-2)(3-x)$

$\large \Rightarrow x^2-2x=-6+2x$

$\large \Rightarrow x^2-2x-2x+6=$ 0

$\large \Rightarrow x^2-4x+6=0$

Yes, It match the Quadratic Equation Format.

∵ $\large ax^{2}+bx+c=0$

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

Solution:

(x – 2)(x + 1) = (x – 1)(x + 3)

$\large \Rightarrow x(x+1)-2(x+1)=x(x+3-1(x+3)$

$\large \Rightarrow x(x+1)-2(x+1)=x(x+3-1(x+3)$

$\large \Rightarrow x(x+1)-2(x+1)=x(x+3)-1(x+3)$

$\large \Rightarrow x^2+x-2x-2=x^2+3x-x-3$

$\large \Rightarrow x^2-x^2+x-2x-3x+x-2+3=0$

$\large \Rightarrow -3x-2+3=0$

$\large \Rightarrow -3x+1=0$

No, Do not match the Quadratic Equation Format.

∵ $\large ax^{2}+bx+c=0$

(iv) (x – 3)(2x +1) = x(x + 5)

Solution:

(x – 3)(2x +1) = x(x + 5)

$\large \Rightarrow x(2x+1)-3(2x+1)=x(x+5)$

$\large \Rightarrow 2x^2+x-6x-3=x^2+5x$

$\large \Rightarrow 2x^2-x^2+x-6x-5x-3=0$

$\large \Rightarrow x^2-10x-3=0$

Yes, It match the Quadratic Equation Format.

∵ $\large ax^{2}+bx+c=0$

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

Solution:

(2x – 1)(x – 3) = (x + 5)(x – 1)

$\large \Rightarrow 2x(x-3)-1(x-3)= x(x-1)+5(x-1)$

$\large \Rightarrow 2x^2-6x-x+3= x^2-x+5x-5$

$\large \Rightarrow 2x^2-x^2-6x-5x-x+x+3+5=0$

$\large \Rightarrow x^2-11x+8=0$

Yes, It match the Quadratic Equation Format.

∵ $\large ax^{2}+bx+c=0$

(vi) $\large x^2+3x+1=(x-2)^2$

Solution:

$\large x^2+3x+1=(x-2)^2$

$\large \Rightarrow x^2+3x+1=x^2+2^2-2.x.2$

$\large \Rightarrow x^2-x^2+3x+4x+1-2^2=0$

$\large \Rightarrow 7x+1-4=0$

$\large \Rightarrow 7x-3=0$

No, Do not match the Quadratic Equation Format.

∵ $\large ax^{2}+bx+c=0$

(vii) $\large (x+2)^3=2x(x^2-1)$

Solution:

$\large (x+2)^3=2x(x^2-1)$

$\large \Rightarrow x^3+3.x^2.2+3.x.2^2+2^3=2x^3-2x$

$\large \Rightarrow x^3+6x^2+12x+2^3=2x^3-2x$

$\large \Rightarrow x^3-2x^3+6x^2+2x+12x+8=0$

$\large \Rightarrow -x^3+6x^2+14x+8=0$

No, Do not match the Quadratic Equation Format.

∵ $\large ax^{2}+bx+c=0$

(viii) $\large x^3-4x^2-x+1=(x-2)^3$

Solution:

$\large x^3-4x^2-x+1=(x-2)^3$

$\large \Rightarrow x^3-4x^2-x+1=x^3-3.x^2.2+3.x.2^2-2^3$

$\large \Rightarrow x^3-x^3-4x^2-x+1=-6x^2+12x-8$ $\large \Rightarrow -4x^2+6x^2-x-12x+1+8=0$

$\large \Rightarrow 2x^2-13x+9=0$

Yes, It match the Quadratic Equation Format.

∵ $\large ax^{2}+bx+c=0$

2. Represent the following situations in the form of quadratic equations :

(i) The area of a rectangular plot is 528 $\large m^2$ . The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

Solution:

Let the length and breadth of the rectangular plot be L and B.

Again, Let breadth be x

A/Q

∵ B=x

∵ L= 2x+1

Given

Area of a rectangular plot=528 $\large m^2$

⇨L X B=528

⇨(2x+1)x=528

⇨ $\large 2x^2$ +x=528

⇨ $\large 2x^2$ +x-528=0

From the above quadratic equation we can find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Solution:

Let the two consecutive positive integers are x, x+1

A/Q

x(x+1)=306

⇨ $\large x^2$ +x=306

⇨ $\large x^2$ +x- 306 = 0

From the above quadratic equation We can 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.

Solution:

Let the present age of Rohan be x

A/Q

Age of Rohan=x

Age of Rohan’s mother=x+26

Age 3 years from now

Rohan=x+3

Rohan’s mother=x+26+3

∵ (x+3)(x+26+3)=360

⇨ $\large x^2$ +26x+3x+3x+78+9=360

⇨ $\large x^2$ +32x+87=360

⇨ $\large x^2$ +32x+87-360=0

⇨ $\large x^2$ +32x-273=0

From the above quadratic equation we can 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.

Solution:

Let the initial speed of train be u km/h

Therefore time required to travel 480 km is

$\large t=\frac{d}{v}$

$\large \Rightarrow t=\frac{480}{u}$ sec

Now V=u-8 km/h

And t=t+3

Again,

$\large d=vt$

$\large \Rightarrow 480=(u-8)t$

$\large \Rightarrow 480=(u-8)(\frac{480}{u}+3)$

$\large \Rightarrow 480=(u-8)(\frac{480+3u}{u})$

$\large \Rightarrow 480u=(u-8)(480+3u)$

$\large \Rightarrow 480u=480u+3u^2-3840-24u$

$\large \Rightarrow 480u-480u=3u^2-3840-24u$

$\large \Rightarrow 3u^2-24u-3840=0$

$\large \Rightarrow u^2-8u-1280=0$

From the above quadratic equation we can find speed of the train.

Tags: Chapter-4 Exercise-4.1 Math- Class 10

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