Solve the equation 6 x - 2 x - 1 = 1 x - 2 by factorisation.

We are given to solve the equation 6 x - 2 x - 1 = 1 x - 2 by factorisation. First, we shall simplify the given equation, convert it into a quadratic equation in its standard form and then, use the method of splitting the middle term for the factorisation. Thus, we have, 6 x - 2 x - 1 = 1 x - 2 ⇒ 6 x - 1 - 2 x x x - 1 = 1 x - 2 ⇒ 6 x - 6 - 2 x x 2 - x = 1 x - 2 ⇒ 4 x - 6 x 2 - x = 1 x - 2 ⇒ 4 x - 6 x - 2 = x 2 - x ⇒ 4 x 2 - 8 x - 6 x + 12 - x 2 + x = 0 ⇒ 3 x 2 - 13 x + 12 = 0 ⇒ 3 x 2 - 9 x - 4 x + 12 = 0 ⇒ 3 x x - 3 - 4 x - 3 = 0 ⇒ x - 3 3 x - 4 = 0 ⇒ x - 3 = 0 or 3 x - 4 = 0 ⇒ x = 3 or x = 4 3 . Therefore, the solutions of the given equation are 3 and 4 3 .

Solve the equation x + 15 = x + 3 by factorisation.

We are given to solve the equation x + 15 = x + 3 by factorisation. First, we shall simplify the given equation, convert it into a quadratic equation in its standard form and then, use the method of splitting the middle term for the factorisation. Thus, we have, x + 15 = x + 3 ⇒ x + 15 2 = x + 3 2 ⇒ x + 15 = x 2 + 6 x + 9 ⇒ x 2 + 6 x + 9 - x - 15 = 0 ⇒ x 2 + 5 x - 6 = 0 ⇒ x 2 + 6 x - x - 6 = 0 ⇒ x x + 6 - 1 x + 6 = 0 ⇒ x + 6 x - 1 = 0 ⇒ x + 6 = 0 or x - 1 = 0 ⇒ x = - 6 or x = 1 . Therefore, the solutions of the given equation are - 6 and 1 .

Solve the equation 3 x 2 - 2 x - 1 = 2 x - 2 by factorisation.

We are given to solve the equation 3 x 2 - 2 x - 1 = 2 x - 2 by factorisation. First, we shall simplify the given equation, convert it into a quadratic equation in its standard form and then, use the method of splitting the middle term for the factorisation. Thus, we have, 3 x 2 - 2 x - 1 = 2 x - 2 ⇒ 3 x 2 - 2 x - 1 2 = 2 x - 2 2 ⇒ 3 x 2 - 2 x - 1 = 4 x 2 - 8 x + 4 ⇒ 4 x 2 - 8 x + 4 - 3 x 2 + 2 x + 1 = 0 ⇒ x 2 - 6 x + 5 = 0 ⇒ x 2 - 5 x - x + 5 = 0 ⇒ x x - 5 - 1 x - 5 = 0 ⇒ x - 5 x - 1 = 0 ⇒ x - 5 = 0 or x - 1 = 0 ⇒ x = 5 or x = 1 . Therefore, the solutions of the given equation are 5 and 1 .

Solve the equation x 3 x + 1 2 = 6 by using formula.

We are given to solve the equation x 3 x + 1 2 = 6 by using the formula. First, we shall simplify the given equation into a quadratic equation in its standard form and then, use the quadratic formula to solve the given equation. Thus, we have, x 3 x + 1 2 = 6 ⇒ 3 x 2 + x 2 - 6 = 0 ⇒ 3 x 2 + x 2 - 6 × 2 = 0 × 2 ⇒ 6 x 2 + x - 12 = 0 We know that in the standard form of quadratic equation a x 2 + b x + c = 0 , where a , b , c are real numbers and a ≠ 0 , the roots are given by x = - b ± b 2 - 4 a c 2 a . Comparing the given equation 6 x 2 + x - 12 = 0 with the standard form of the equation, we get, a = 6 , b = 1 and c = - 12 . So, the roots are x = - 1 ± 1 2 - 4 × 6 × - 12 2 × 6 ⇒ x = - 1 ± 1 + 288 12 ⇒ x = - 1 ± 289 12 ⇒ x = - 1 ± 17 12 ⇒ x = - 1 + 17 12 , - 1 - 17 12 ⇒ x = 16 12 , - 18 12 ⇒ x = 4 3 , - 3 2 Therefore, the roots of the given quadratic equation are 4 3 and - 3 2 .

E M B I B E

Understanding ICSE Mathematics Class 10>Algebra>Chapter 5 - Quadratic Equations in One Variable>Chapter Test

M L Aggarwal Solutions for Exercise 7: Chapter Test

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Exercise 7: Chapter Test

Attempt the practice questions from Exercise 7: Chapter Test with hints and solutions to strengthen your understanding. Understanding ICSE Mathematics Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Exercise 7: Chapter Test with Hints & Solutions

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 5 - Quadratic Equations in One Variable>Chapter Test>Q 1

Solve the equation $x^{2} + 6 x - 16 = 0$ by factorisation.

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 5 - Quadratic Equations in One Variable>Chapter Test>Q 3

Solve the equation $x (x + 1) + (x + 2) (x + 3) = 42$ by factorisation.

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 5 - Quadratic Equations in One Variable>Chapter Test>Q 3

Solve the equation $\frac{6}{x} - \frac{2}{x - 1} = \frac{1}{x - 2}$ by factorisation.

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 5 - Quadratic Equations in One Variable>Chapter Test>Q 4

Solve the equation $\sqrt{x + 15} = x + 3$ by factorisation.

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 5 - Quadratic Equations in One Variable>Chapter Test>Q 4

Solve the equation $\sqrt{3 x^{2} - 2 x - 1} = 2 x - 2$ by factorisation.

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 5 - Quadratic Equations in One Variable>Chapter Test>Q 5

Solve the equation $2 x^{2} - 3 x - 1 = 0$ by using the formula.

MEDIUM

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 5 - Quadratic Equations in One Variable>Chapter Test>Q 5

Solve the equation $x (3 x + \frac{1}{2}) = 6$ by using formula.

HARD

10th ICSE

IMPORTANT

Understanding ICSE Mathematics Class 10>Chapter 5 - Quadratic Equations in One Variable>Chapter Test>Q 23

A man spent $₹ 2800$ on buying a number of plants priced at $₹ x$ each. Because of the number involved, the supplier reduced the price of each plant by one rupee. The man finally paid $₹ 2730$ and received $10$ more plants. Find $x$ .

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