In each of the following, determine whether the given values are solutions of the given equation or not:
(i) X²-3X+2=0,X=2,X=1
(ii) X²+X+1=0,X=1
(iii) X²-3√3X+6=0,X=√3,X=2√3
(iv) X+1/X=13/6,X=5/6,X=-2√3
(v) 2X²-X+9=X²+3,X=2,X=3
(vi) X²-√2X-4=0,X=-√2,X=-2√2
(vii)a²x²-3abx+2b²=0,x=a/b,x=b/a
Answers
(i) Yes, (ii) No, (iii) Yes, (iv) No, (v) Yes, (vi) No, (vii) No.
Step-by-step explanation:
(i) X² - 3X + 2 = 0
Factorizing, we get
(X - 1)(X - 2) = 0
So, the roots are X = 1 and X = 2.
(ii) X² + X + 1 = 0
Can not be factorized.
No real root.
(iii) X² - 3√3X + 6 = 0
Factorizing we get,
X² - √3X - 2√3X + 6 = 0
⇒ (X - √3)(X - 2√3) = 0
So, the roots are X = √3 and X = 2√3.
(iv) X + 1/X = 13/6
⇒ 6X² - 13X + 6 = 0
Factorizing we get,
6X² - 4X - 9X + 6 = 0
⇒ (3X - 2)(2X - 3) = 0
So, the roots are X = 2/3 or X = 3/2.
(v) 2X² - X + 9 = X² + 3
⇒ X² - X + 6 = 0
Factorizing we get,
X² - 3X - 2X + 6 = 0
⇒ (X - 3)(X - 2) = 0
So, the roots are X = 3 and X = 2.
(vi) X² - √2X - 4 = 0
Factorizing we get,
X² + 2√2X - √2X - 4 = 0
⇒ (X + 2√2)(X - √2) = 0
So, the roots are X = - 2√2 and X = √2.
(vii) a²x² - 3abx + 2b² = 0
Factorizing we get,
a²x² - abx - 2abx + 2b² = 0
⇒ (ax - b)(ax - 2b) = 0
So, the roots are, x = b/a and x = 2b/a. (Answer)