Question

Question 3 Find the roots of the following equations:
(i) x - 1/x = 3, x ≠ 0
(ii) 1/(x+4) - 1/(x-7) = 11/30, x ≠ -4,7

Class 10 - Math - Quadratic Equations Page 88

Answer 1

(i) x-1/x = 3
⇒ x2 - 3x -1 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -3 and c = -1
By using quadratic formula, we get
x = -b±√b2 - 4ac/2a
⇒ x = 3±√9+4/2
⇒ x = 3±√13/2
∴ x = 3+√13/2 or x = 3-√13/2

(ii) 1/x+4 - 1/x-7 = 11/30
⇒ x-7-x-4/(x+4)(x-7) = 11/30
⇒ -11/(x+4)(x-7) = 11/30
⇒ (x+4)(x-7) = -30
⇒ x2 - 3x - 28 = 30
⇒ x2 - 3x + 2 = 0
⇒ x2 - 2x - x + 2 = 0
⇒ x(x - 2) - 1(x - 2) = 0
⇒ (x - 2)(x - 1) = 0
⇒ x = 1 or 2

Answer 2

Method to solve a quadratic equation by using quadratic formula.

Step 1. Write a given quadratic equation in standard form.ax²+bx+c=0 ,
Step 2.Find the value of a, b and c by comparing the given equation with ax²+bx+c=0 ,
Step 3. Put the values of a, b and c  in quadratic formula i.e X=( -b +_ √ b²-4ac) /2a
By simplifying the  above  formula we get two roots…
This method is also  called Sridharacharya’s rule.
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By using quadratic formula

(i) x-1/x = 3
⇒ x² – 3x -1 = 0

On comparing this equation with ax² + bx + c = 0, we get
a = 1, b = -3 and c = -1

By using quadratic formula, we get
x = –b±√b2 – 4ac/2a

⇒ x = (3±√9+4 )/2
⇒ x = (3±√13 )/2

∴ x = (3+√13 )/2 or x = (3-√13 )/2

(ii) 1/x+4 – 1/x-7 = 11/30

⇒ x-7-x-4/(x+4)(x-7) = 11/30
⇒ -11/(x+4)(x-7) = 11/30

⇒ (x+4)(x-7) = -30
⇒ x² – 3x – 28 = 30

⇒ x² – 3x + 2 = 0                     ( by using factorisation method)
⇒ x² – 2x – x + 2 = 0
⇒ x(x – 2) – 1(x – 2) = 0
⇒ (x – 2)(x – 1) = 0

⇒ x = 1 or 2

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Hope this will help you...