Quadratic Equations | Exercise 4.1

Question 1

  1. Check whether the following are quadratic equations :

(i) (x+1)2=2(x−3)(x + 1)^2 = 2(x - 3)

(ii) x2−2x=(−2)(3−x)x^2 - 2x = (-2)(3 - x)

(iii) (x−2)(x+1)=(x−1)(x+3)(x - 2)(x + 1) = (x - 1)(x + 3)

(iv) (x−3)(2x+1)=x(x+5)(x - 3)(2x + 1) = x(x + 5)

(v) (2x−1)(x−3)=(x+5)(x−1)(2x - 1)(x - 3) = (x + 5)(x - 1)

(vi) x2+3x+1=(x−2)2x^2 + 3x + 1 = (x - 2)^2

(vii) (x+2)3=2x(x2−1)(x + 2)^3 = 2x(x^2 - 1)

(viii) x3−4x2−x+1=(x−2)3x^3 - 4x^2 - x + 1 = (x - 2)^3

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Solution
Understand the Question
  • A quadratic equation in variable xx is an equation of the standard form ax2+bx+c=0ax^2 + bx + c = 0, where a,b,ca, b, c are real numbers and a≠0a \neq 0.
  • To check whether a given equation is quadratic:
    1. Expand and simplify both sides completely.
    2. Move all terms to one side so the equation is in the form P(x)=0P(x) = 0.
    3. If the highest power (degree) of xx is 22 and the coefficient of x2x^2 is non-zero, it is a quadratic equation.

(i) Check whether (x+1)2=2(x−3)(x + 1)^2 = 2(x - 3) is a quadratic equation.

Step 1 · Simplify and Check Degree

Expand both sides

(x+1)2=2(x−3)x2+2x+1=2x−6x2+2x−2x+1+6=0x2+7=0\begin{aligned} (x + 1)^2 &= 2(x - 3) \\ x^2 + 2x + 1 &= 2x - 6 \\ x^2 + 2x - 2x + 1 + 6 &= 0 \\ x^2 + 7 &= 0 \end{aligned}

It is of the form ax2+bx+c=0ax^2 + bx + c = 0, where a=1≠0a = 1 \neq 0, b=0b = 0, and c=7c = 7.

Answer

(i) Yes, it is a quadratic equation.

(ii) Check whether x2−2x=(−2)(3−x)x^2 - 2x = (-2)(3 - x) is a quadratic equation.

Step 1 · Simplify and Check Degree

Expand both sides

x2−2x=(−2)(3−x)x2−2x=−6+2xx2−2x−2x+6=0x2−4x+6=0\begin{aligned} x^2 - 2x &= (-2)(3 - x) \\ x^2 - 2x &= -6 + 2x \\ x^2 - 2x - 2x + 6 &= 0 \\ x^2 - 4x + 6 &= 0 \end{aligned}

It is of the form ax2+bx+c=0ax^2 + bx + c = 0, where a=1≠0a = 1 \neq 0, b=−4b = -4, and c=6c = 6.

Answer

(ii) Yes, it is a quadratic equation.

(iii) Check whether (x−2)(x+1)=(x−1)(x+3)(x - 2)(x + 1) = (x - 1)(x + 3) is a quadratic equation.

Step 1 · Simplify and Check Degree

Expand both sides

(x−2)(x+1)=(x−1)(x+3)x2+x−2x−2=x2+3x−x−3x2−x−2=x2+2x−3x2−x2−x−2x−2+3=0−3x+1=0\begin{aligned} (x - 2)(x + 1) &= (x - 1)(x + 3) \\ x^2 + x - 2x - 2 &= x^2 + 3x - x - 3 \\ x^2 - x - 2 &= x^2 + 2x - 3 \\ x^2 - x^2 - x - 2x - 2 + 3 &= 0 \\ -3x + 1 &= 0 \end{aligned}

The degree of the equation is 11, not 22.

Answer

(iii) No, it is not a quadratic equation.

(iv) Check whether (x−3)(2x+1)=x(x+5)(x - 3)(2x + 1) = x(x + 5) is a quadratic equation.

Step 1 · Simplify and Check Degree

Expand both sides

(x−3)(2x+1)=x(x+5)2x2+x−6x−3=x2+5x2x2−5x−3=x2+5x2x2−x2−5x−5x−3=0x2−10x−3=0\begin{aligned} (x - 3)(2x + 1) &= x(x + 5) \\ 2x^2 + x - 6x - 3 &= x^2 + 5x \\ 2x^2 - 5x - 3 &= x^2 + 5x \\ 2x^2 - x^2 - 5x - 5x - 3 &= 0 \\ x^2 - 10x - 3 &= 0 \end{aligned}

It is of the form ax2+bx+c=0ax^2 + bx + c = 0, where a=1≠0a = 1 \neq 0, b=−10b = -10, and c=−3c = -3.

Answer

(iv) Yes, it is a quadratic equation.

(v) Check whether (2x−1)(x−3)=(x+5)(x−1)(2x - 1)(x - 3) = (x + 5)(x - 1) is a quadratic equation.

Step 1 · Simplify and Check Degree

Expand both sides

(2x−1)(x−3)=(x+5)(x−1)2x2−6x−x+3=x2−x+5x−52x2−7x+3=x2+4x−52x2−x2−7x−4x+3+5=0x2−11x+8=0\begin{aligned} (2x - 1)(x - 3) &= (x + 5)(x - 1) \\ 2x^2 - 6x - x + 3 &= x^2 - x + 5x - 5 \\ 2x^2 - 7x + 3 &= x^2 + 4x - 5 \\ 2x^2 - x^2 - 7x - 4x + 3 + 5 &= 0 \\ x^2 - 11x + 8 &= 0 \end{aligned}

It is of the form ax2+bx+c=0ax^2 + bx + c = 0, where a=1≠0a = 1 \neq 0, b=−11b = -11, and c=8c = 8.

Answer

(v) Yes, it is a quadratic equation.

(vi) Check whether x2+3x+1=(x−2)2x^2 + 3x + 1 = (x - 2)^2 is a quadratic equation.

Step 1 · Simplify and Check Degree

Expand both sides

x2+3x+1=(x−2)2x2+3x+1=x2−4x+4x2−x2+3x+4x+1−4=07x−3=0\begin{aligned} x^2 + 3x + 1 &= (x - 2)^2 \\ x^2 + 3x + 1 &= x^2 - 4x + 4 \\ x^2 - x^2 + 3x + 4x + 1 - 4 &= 0 \\ 7x - 3 &= 0 \end{aligned}

The degree of the equation is 11, not 22.

Answer

(vi) No, it is not a quadratic equation.

(vii) Check whether (x+2)3=2x(x2−1)(x + 2)^3 = 2x(x^2 - 1) is a quadratic equation.

Step 1 · Simplify and Check Degree

Using the identity (a+b)3=a3+3a2b+3ab2+b3(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

(x+2)3=2x(x2−1)x3+3(x2)(2)+3(x)(22)+23=2x3−2xx3+6x2+12x+8=2x3−2xx3−2x3+6x2+12x+2x+8=0−x3+6x2+14x+8=0\begin{aligned} (x + 2)^3 &= 2x(x^2 - 1) \\ x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3 &= 2x^3 - 2x \\ x^3 + 6x^2 + 12x + 8 &= 2x^3 - 2x \\ x^3 - 2x^3 + 6x^2 + 12x + 2x + 8 &= 0 \\ -x^3 + 6x^2 + 14x + 8 &= 0 \end{aligned}

The highest power of xx is 33, which means the degree is 33 (a cubic equation).

Answer

(vii) No, it is not a quadratic equation.

(viii) Check whether x3−4x2−x+1=(x−2)3x^3 - 4x^2 - x + 1 = (x - 2)^3 is a quadratic equation.

Step 1 · Simplify and Check Degree

Using the identity (a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

x3−4x2−x+1=(x−2)3x3−4x2−x+1=x3−3(x2)(2)+3(x)(22)−23x3−4x2−x+1=x3−6x2+12x−8x3−x3−4x2+6x2−x−12x+1+8=02x2−13x+9=0\begin{aligned} x^3 - 4x^2 - x + 1 &= (x - 2)^3 \\ x^3 - 4x^2 - x + 1 &= x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 \\ x^3 - 4x^2 - x + 1 &= x^3 - 6x^2 + 12x - 8 \\ x^3 - x^3 - 4x^2 + 6x^2 - x - 12x + 1 + 8 &= 0 \\ 2x^2 - 13x + 9 &= 0 \end{aligned}

It is of the form ax2+bx+c=0ax^2 + bx + c = 0, where a=2≠0a = 2 \neq 0, b=−13b = -13, and c=9c = 9.

Answer

(viii) Yes, it is a quadratic equation.

Common Mistakes
  • Judging Without Simplifying: Assuming an expression like (x−2)(x+1)=(x−1)(x+3)(x-2)(x+1) = (x-1)(x+3) is quadratic because of the product of xx terms, even though the x2x^2 terms cancel out.
  • Prematurely Rejecting Cubic Terms: Assuming an equation with cubic terms like x3−4x2−x+1=(x−2)3x^3 - 4x^2 - x + 1 = (x - 2)^3 cannot be quadratic without expanding to see that the x3x^3 terms cancel out on both sides.
  • Algebraic Identity Errors: Making sign mistakes when expanding cubic expressions, such as (a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.

More questions in Exercise 4.1

Q1
  1. Check whether the following are quadratic equations :

(i) (x+1)2=2(x−3)(x + 1)^2 = 2(x - 3)

(ii) x2−2x=(−2)(3−x)x^2 - 2x = (-2)(3 - x)

(iii) (x−2)(x+1)=(x−1)(x+3)(x - 2)(x + 1) = (x - 1)(x + 3)

(iv) (x−3)(2x+1)=x(x+5)(x - 3)(2x + 1) = x(x + 5)

(v) (2x−1)(x−3)=(x+5)(x−1)(2x - 1)(x - 3) = (x + 5)(x - 1)

(vi) x2+3x+1=(x−2)2x^2 + 3x + 1 = (x - 2)^2

(vii) (x+2)3=2x(x2−1)(x + 2)^3 = 2x(x^2 - 1)

(viii) x3−4x2−x+1=(x−2)3x^3 - 4x^2 - x + 1 = (x - 2)^3

Q2
  1. Represent the following situations in the form of quadratic equations :

(i) The area of a rectangular plot is 528 m2528\text{ m}^2. 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.

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