Exercise 4.1 অনুশীলনী ৪.১
১. তলৰ বোৰ দ্বিঘাত সমীকৰণ হয়নে পৰীক্ষা কৰা :
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
সমাধান:
Solution:
(i) (x + 1)² = 2(x – 3)
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
The above equation is in the form of ax² + bx + c = 0
Therefore, the given equation is a quadratic equation.
For Assamese Medium
দিয়া আছে
(x + 1)² = 2(x – 3)
এতিয়া (a+b)²= a²+2ab+b² সূত্ৰ ব্যৱহাৰ কৰি পাওঁ
⇒ x² + 2x + 1 = 2x – 6
⇒ x² + 7 = 0
ওপৰোক্ত সমীকৰণটো ax² + bx + c = 0 ৰ ৰূপত আছে।
সেয়েহে, প্ৰদত্ত সমীকৰণটো হৈছে এক দ্বিঘাত সমীকৰণ।
(ii) x² – 2x = (–2) (3 – x)
Given, x² – 2x = (–2) (3 – x)
⇒ x² – 2x = -6 + 2x
⇒ x²– 4x + 6 = 0
The above equation is in the form of ax² + bx + c = 0
Therefore, the given equation is a quadratic equation.
For Assamese Medium:
দিয়া আছে, x² – 2x = (–2) (3 – x)
⇒ x² – 2x = -6 + 2x
⇒ x²– 4x + 6 = 0
ওপৰোক্ত সমীকৰণটো ax² + bx + c = 0 ৰ ৰূপত আছে।
সেয়েহে, প্ৰদত্ত সমীকৰণটো হৈছে এক দ্বিঘাত সমীকৰণ।
(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3)
By multiplication,
⇒ x2– x – 2 = x2+ 2x – 3
⇒ 3x – 1 = 0
The above equation is not in the form of ax2 + bx + c = 0
Therefore, the given equation is not a quadratic equation.
(iv) Given, (x – 3)(2x +1) = x(x + 5)
By multiplication,
⇒ 2x2– 5x – 3 = x2+ 5x
⇒ x2– 10x – 3 = 0
The above equation is in the form of ax2 + bx + c = 0
Therefore, the given equation is a quadratic equation.
(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1)
By multiplication,
⇒ 2x2– 7x + 3 = x2+ 4x – 5
⇒ x2– 11x + 8 = 0
The above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is a 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
The above equation is not in the form of ax2 + bx + c = 0
Therefore, the given equation is not a quadratic equation.
(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
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
The above equation is in the form of ax2 + bx + c = 0
Therefore, the given equation is a quadratic equation.