krit.club logo

Algebra - Linear and Quadratic Equations

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Solving Linear Equations: Isolating the variable by performing inverse operations on both sides of the equation.

•

Rearranging Formulae: Changing the subject of a formula using algebraic manipulation.

•

Expanding Brackets: Using the distributive law and the FOIL method for binomials.

•

Factorising Quadratics: Finding two binomial factors for expressions in the form ax2+bx+cax^2 + bx + c.

•

Difference of Two Squares: Recognising and factorising expressions like a2−b2a^2 - b^2.

•

Solving Quadratic Equations: Finding the values of xx that satisfy the equation using factorisation or the quadratic formula.

•

The Null Factor Law: If aimesb=0a imes b = 0, then a=0a = 0 or b=0b = 0.

📐Formulae

ax+b=cax + b = c

ax2+bx+c=0ax^2 + bx + c = 0

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

(a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2

💡Examples

Problem 1:

Solve the linear equation: 4(x−3)=2x+104(x - 3) = 2x + 10

Solution:

x=11x = 11

Explanation:

First, expand the bracket: 4x−12=2x+104x - 12 = 2x + 10. Subtract 2x2x from both sides: 2x−12=102x - 12 = 10. Add 12 to both sides: 2x=222x = 22. Divide by 2: x=11x = 11.

Problem 2:

Factorise and solve: x2−5x−14=0x^2 - 5x - 14 = 0

Solution:

x=7x = 7 or x=−2x = -2

Explanation:

Find two numbers that multiply to −14-14 and add to −5-5. These are −7-7 and +2+2. Factorised form: (x−7)(x+2)=0(x - 7)(x + 2) = 0. Set each factor to zero: x−7=0⇒x=7x - 7 = 0 \Rightarrow x = 7; x+2=0⇒x=−2x + 2 = 0 \Rightarrow x = -2.

Problem 3:

Solve 2x2+7x+1=02x^2 + 7x + 1 = 0 using the quadratic formula. Give answers to 2 decimal places.

Solution:

x≈−0.15x \approx -0.15 or x≈−3.35x \approx -3.35

Explanation:

Identify a=2,b=7,c=1a=2, b=7, c=1. Substitute into the formula: x=−7±72−4(2)(1)2(2)x = \frac{-7 \pm \sqrt{7^2 - 4(2)(1)}}{2(2)}. This simplifies to x=−7±414x = \frac{-7 \pm \sqrt{41}}{4}. Calculating the two values: x=−7+6.4034=−0.149...x = \frac{-7 + 6.403}{4} = -0.149... and x=−7−6.4034=−3.350...x = \frac{-7 - 6.403}{4} = -3.350....

Problem 4:

Rearrange the formula to make hh the subject: V=13πr2hV = \frac{1}{3}\pi r^2 h

Solution:

h=3Vπr2h = \frac{3V}{\pi r^2}

Explanation:

Multiply both sides by 3 to remove the fraction: 3V=πr2h3V = \pi r^2 h. Divide both sides by πr2\pi r^2 to isolate hh: h=3Vπr2h = \frac{3V}{\pi r^2}.