Copy and complete: x 2 - □ x + 15 ≡ ( x - 3 ) ( x - □ )

Given that, x 2 - □ x + 15 ≡ ( x - 3 ) ( x - □ ) . Let the unknown boxes are filled with a & b . i.e., x 2 - a x + 15 ≡ ( x - 3 ) ( x - b ) ⇒ x 2 - a x + 15 ≡ x 2 - 3 x - b x + 3 b = x 2 - 3 + b x + 3 b Equating corresponding coefficients, we get a = 3 + b and 15 = 3 b ⇒ b = 5 and a = 3 + 5 = 8 Therefore, given expression is x 2 - 8 x + 15 ≡ ( x - 3 ) ( x - 5 )

Copy and complete: x 2 □ □ x + 20 ≡ ( x + 4 ) ( x □ □ )

Given that, x 2 □ □ x + 20 ≡ ( x + 4 ) ( x □ □ ) . Let the unknown boxes are filled with + a & + b . i.e., x 2 + a x + 20 ≡ ( x + 4 ) ( x + b ) ⇒ x 2 + a x + 20 ≡ x 2 + 4 x + b x + 4 b = x 2 + 4 + b x + 4 b Equating corresponding coefficients, we get a = 4 + b and 20 = 4 b ⇒ b = 5 and a = 4 + 5 = 9 Therefore, given expression is x 2 + 9 x + 20 = x + 4 x + 5

Copy and complete: x 2 - 8 x □ □ ≡ ( x + □ ) ( x - 11 )

Given that, x 2 - 8 x □ □ ≡ ( x + □ ) ( x - 11 ) . Let the unknown boxes are filled with + a & b . i.e., x 2 - 8 x + a ≡ ( x + b ) ( x - 11 ) ⇒ x 2 - 8 x + a ≡ x 2 + b x - 11 x - 11 b = x 2 + b - 11 x - 11 b Equating corresponding coefficients, we get a = - 11 b and - 8 = b - 11 ⇒ b = - 3 and a = - 11 3 = - 33 Therefore, given expression is x 2 - 8 x - 33 = ( x + 3 ) ( x - 11 )

Copy and complete: x 2 □ 11 x + □ ≡ ( x □ 5 ) ( x - □ )

Given that, x 2 □ 11 x + □ &equiv; ( x □ 5 ) ( x - □ ) . 11 in LHS can be ± 11 11 is written as 5 + 6 - 11 can be written 5 - 16 For, 5 , -16 as factors, we get - 80 as constant term on RHS, which is not acceptable as the given constant on LHS is positive. Therefore, given expression can be written as x 2 - 11 x + 30 = ( x - 5 ) ( x - 6 ) .

E M B I B E

MYP Mathematics A concept based approach 4&5 Standard>Patterns>Chapter 17 - Back to the Beginning (Using Patterns to Work Backwards)>Mixed practice

Rose Harrison and Clara Huizink Solutions for Exercise 21: Mixed practice

Author:Rose Harrison & Clara Huizink

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison and Clara Huizink Solutions for Exercise 21: Mixed practice

Attempt the practice questions from Exercise 21: Mixed practice with hints and solutions to strengthen your understanding. MYP Mathematics A concept based approach 4&5 Standard solutions are prepared by Experienced Embibe Experts.

Questions from Rose Harrison and Clara Huizink Solutions for Exercise 21: Mixed practice with Hints & Solutions

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 17 - Back to the Beginning (Using Patterns to Work Backwards)>Mixed practice>Q 6

Factorize: $3 g^{2} - 23 g + 14$

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 17 - Back to the Beginning (Using Patterns to Work Backwards)>Mixed practice>Q 6

Factorize: $16 h^{2} - 1$

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 17 - Back to the Beginning (Using Patterns to Work Backwards)>Mixed practice>Q 8

Copy and complete:

$x^{2} - □ x + 15 \equiv (x - 3) (x - □)$

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 17 - Back to the Beginning (Using Patterns to Work Backwards)>Mixed practice>Q 8

Copy and complete:

$x^{2} □ □ x + 20 \equiv (x + 4) (x □ □)$

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 17 - Back to the Beginning (Using Patterns to Work Backwards)>Mixed practice>Q 8

Copy and complete:

$x^{2} - 8 x □ □ \equiv (x + □) (x - 11)$

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 17 - Back to the Beginning (Using Patterns to Work Backwards)>Mixed practice>Q 8

Copy and complete:

$x^{2} □ 11 x + □ \equiv (x □ 5) (x - □)$

MEDIUM

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 17 - Back to the Beginning (Using Patterns to Work Backwards)>Mixed practice>Q 9

Dagmar thinks of an even number, squares it, and subtracts $1$ . By expressing his original number in the form $2 n$ , or otherwise, show that the result of this process can be written as the product of two consecutive odd integers.

MEDIUM

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 17 - Back to the Beginning (Using Patterns to Work Backwards)>Mixed practice>Q 11

A rectangular sports pitch has a border of $n$ meters on each side. The total area occupied by the pitch and its border is $4 n^{2} + 28 n + 45$ .

Question Image