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Algebra - Linear and Quadratic Equations

Grade 10IGCSE

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

🔑Concepts

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Solving linear equations by isolating the variable using inverse operations.

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Solving simultaneous linear equations using the elimination method or substitution method.

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Rearranging literal equations (changing the subject of a formula).

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Solving quadratic equations of the form ax2+bx+c=0ax^2 + bx + c = 0 by factorisation.

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Using the Quadratic Formula for equations that cannot be easily factorised.

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Completing the square to find the vertex of a parabola and solve equations.

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Understanding the nature of roots using the Discriminant (Δ=b2−4ac\Delta = b^2 - 4ac).

📐Formulae

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

Discriminant: Δ=b2−4ac\Delta = b^2 - 4ac

Completing the Square form: (x+b2)2−(b2)2+c=0(x + \frac{b}{2})^2 - (\frac{b}{2})^2 + c = 0

Standard form of a Quadratic: ax2+bx+c=0ax^2 + bx + c = 0

Slope-intercept form of a linear equation: y=mx+cy = mx + c

💡Examples

Problem 1:

Solve the simultaneous equations: 3x+2y=73x + 2y = 7 and x−y=4x - y = 4.

Solution:

x=3,y=−1x = 3, y = -1

Explanation:

Multiply the second equation by 2 to get 2x−2y=82x - 2y = 8. Add this to the first equation: (3x+2x)+(2y−2y)=7+8(3x + 2x) + (2y - 2y) = 7 + 8, which simplifies to 5x=155x = 15, so x=3x = 3. Substitute x=3x = 3 into the second equation: 3−y=4⇒y=−13 - y = 4 \Rightarrow y = -1.

Problem 2:

Solve 2x2−5x−3=02x^2 - 5x - 3 = 0 using factorisation.

Solution:

x=3x = 3 or x=−0.5x = -0.5

Explanation:

Find two numbers that multiply to 2×(−3)=−62 \times (-3) = -6 and add to −5-5. These are −6-6 and 11. Split the middle term: 2x2−6x+x−3=02x^2 - 6x + x - 3 = 0. Factor by grouping: 2x(x−3)+1(x−3)=02x(x - 3) + 1(x - 3) = 0. This gives (2x+1)(x−3)=0(2x + 1)(x - 3) = 0. Solving for xx gives x=−1/2x = -1/2 and x=3x = 3.

Problem 3:

Solve x2+4x−1=0x^2 + 4x - 1 = 0 by completing the square.

Solution:

x=−2±5x = -2 \pm \sqrt{5}

Explanation:

Move the constant: x2+4x=1x^2 + 4x = 1. Add (b/2)2=(4/2)2=4(b/2)^2 = (4/2)^2 = 4 to both sides: x2+4x+4=1+4x^2 + 4x + 4 = 1 + 4. Write as a perfect square: (x+2)2=5(x + 2)^2 = 5. Take the square root of both sides: x+2=±5x + 2 = \pm \sqrt{5}. Finally, x=−2±5x = -2 \pm \sqrt{5}.

Problem 4:

Determine the nature of the roots for 3x2+2x+5=03x^2 + 2x + 5 = 0.

Solution:

No real roots.

Explanation:

Calculate the discriminant Δ=b2−4ac\Delta = b^2 - 4ac. Here a=3,b=2,c=5a=3, b=2, c=5. Δ=(2)2−4(3)(5)=4−60=−56\Delta = (2)^2 - 4(3)(5) = 4 - 60 = -56. Since Δ<0\Delta < 0, the equation has no real roots (only imaginary roots).