krit.club logo

Algebra and Graphs - Linear and quadratic equations

Grade 11IGCSE

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.

•

Quadratic Equation Methods: Solving ax2+bx+c=0ax^2 + bx + c = 0 using Factorisation, the Quadratic Formula, or Completing the Square.

•

Simultaneous Equations: Finding the point of intersection between two lines (Linear/Linear) or a line and a curve (Linear/Quadratic) using Elimination or Substitution.

•

Graphical Interpretation: Identifying the gradient (m) and y-intercept (c) for linear graphs, and roots (x-intercepts) and vertices for quadratic graphs.

•

The Discriminant: Using b2−4acb^2 - 4ac to determine the number of real roots (Two real roots if >0>0, one real root if =0=0, no real roots if <0<0).

📐Formulae

y=mx+cy = mx + c (Gradient-intercept form)

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} (Gradient formula)

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

(x+p)2+q=0(x + p)^2 + q = 0 (Completed Square form)

y=a(x−h)2+ky = a(x - h)^2 + k (Vertex form of a quadratic graph)

💡Examples

Problem 1:

Solve the quadratic equation 2x2−7x+3=02x^2 - 7x + 3 = 0 using the quadratic formula.

Solution:

x=−(−7)±(−7)2−4(2)(3)2(2)=7±49−244=7±254x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)} = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4}. Therefore, x=7+54=3x = \frac{7+5}{4} = 3 or x=7−54=0.5x = \frac{7-5}{4} = 0.5.

Explanation:

Identify a=2,b=−7,c=3a=2, b=-7, c=3. Substitute these into the quadratic formula. Simplify the discriminant (sqrt25=5\\sqrt{25} = 5) and solve for both the plus and minus cases.

Problem 2:

Solve the simultaneous equations: y=2x−3y = 2x - 3 and y=x2−4x+5y = x^2 - 4x + 5.

Solution:

2x−3=x2−4x+5⇒x2−6x+8=02x - 3 = x^2 - 4x + 5 \Rightarrow x^2 - 6x + 8 = 0. Factorising gives (x−2)(x−4)=0(x-2)(x-4)=0, so x=2x=2 or x=4x=4. When x=2,y=2(2)−3=1x=2, y=2(2)-3=1. When x=4,y=2(4)−3=5x=4, y=2(4)-3=5. Solutions: (2,1)(2, 1) and (4,5)(4, 5).

Explanation:

Since both equations are equal to yy, set them equal to each other to form a single quadratic equation. Solve for xx, then substitute xx back into the linear equation to find the corresponding yy values.

Problem 3:

Find the coordinates of the turning point (vertex) of the graph y=x2−6x+10y = x^2 - 6x + 10 by completing the square.

Solution:

y=(x−3)2−(−3)2+10=(x−3)2−9+10=(x−3)2+1y = (x - 3)^2 - (-3)^2 + 10 = (x - 3)^2 - 9 + 10 = (x - 3)^2 + 1. The vertex is (3,1)(3, 1).

Explanation:

Halve the coefficient of xx (which is −6-6) to get −3-3 for the bracket (x−3)2(x-3)^2. Subtract the square of that number and add the constant. In the form (x−h)2+k(x-h)^2 + k, the vertex is (h,k)(h, k).