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Maths241Updated 8 Sept 20265 pages

Master the Nat 5 Maths Quadratic Formula

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Amilie du Toit@amiliedutoit_uajk
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The quadratic formula is a powerful tool for solving quadratic equations when they can't be easily factored. This formula works for any quadratic equation in the form ax² + bx + c = 0, giving you the exact values where the equation equals zero.
Nat 5 Maths - Quadratic formula – page 1

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Using the Quadratic Formula

The quadratic formula is: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a} where a is the coefficient of x², b is the coefficient of x, and c is the constant term.

Let's see how it works with an example: x2+6x+2=0x^2+6x+2=0. We identify a=1, b=6, and c=2, then substitute into the formula. After calculating the discriminant (b2−4ac=36−8=28b^2-4ac = 36-8 = 28), we get two solutions: x=−6+282≈−0.35x = \frac{-6 + \sqrt{28}}{2} ≈ -0.35 or x=−6−282≈−5.65x = \frac{-6 - \sqrt{28}}{2} ≈ -5.65.

For another example, 3x2−10x+2=03x^2-10x+2=0, we have a=3, b=-10, and c=2. Working through the formula gives us x≈3.12x ≈ 3.12 or x≈0.21x ≈ 0.21.

Remember: The ± sign in the formula means you'll always get two possible solutions (unless the discriminant equals zero, which gives one solution).

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Solving Word Problems with Quadratics

Real-world problems often require quadratic equations. Let's tackle one: a rectangular playground is 10m longer than it is wide with an area of 1400m².

To solve this, we need to set up a quadratic equation. If the width is w, then the length is w+10w+10. The area formula gives us: 1400 = w+10$$w, which expands to w² + 10w - 1400 = 0.

Using the quadratic formula with a=1, b=10, c=-1400: w=−10±100+56002w = \frac{-10 \pm \sqrt{100+5600}}{2}. This gives us w ≈ 32.75m (we reject the negative solution as width can't be negative). Therefore, the length is 32.75 + 10 = 42.75m.

Pro tip: Always check if your answer makes sense in the context of the problem. For example, measurements like width can't be negative!

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Finding Quadratic Equations from Points

You can determine the equation of a quadratic function if you know some points on the graph. For the form y = kx², you only need one non-zero point.

For example, if the graph passes through (0,0) and (1,3), we can find k by substituting the second point: 3 = k(1)², so k = 3. The equation is y = 3x².

Similarly, for a graph through (0,0) and 2,−82,-8, we substitute to get -8 = k(2)², so k = -2. The equation becomes y = -2x².

Quick check: You can verify your equation by testing it with the original points. If y = 3x² and x = 1, then y should equal 3.

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Understanding the Form y = x−px-p² + q

The form y = x−px-p² + q helps us understand how a parabola is positioned on a graph. The value p shifts the graph horizontally, and q shifts it vertically.

If we know that p = 2 and q = 3, the equation is y = x−2x-2² + 3. This means the parabola is shifted 2 units right and 3 units up from the standard position.

For a graph where p = -2 and q = -1, the equation becomes y = x+2x+2² - 1. This parabola is shifted 2 units left and 1 unit down from the standard position.

Visual aid: Think of p and q as giving the coordinates of the turning point (p,q) - this is where the parabola reaches its minimum or maximum value.

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Finding the Equation of a Parabola hxx

When given specific information about a parabola in the form hxx = x−px-p² + q, we can determine its equation.

If a parabola's axis of symmetry is at x = -2 and it passes through (3,0) and −1,0-1,0, then p = -2 (the axis of symmetry). To find q, we substitute one of the points: 0 = 3−(−2)3-(-2)² + q, which gives us 0 = 25 + q, so q = -25.

However, we should verify with the other point: 0 = −1−(−2)-1-(-2)² + q means 0 = 1 + q, so q = -1. Since we get different values for q, we need to double-check our working.

Important: The axis of symmetry for a parabola in the form x−px-p² + q is always x = p, which is midway between the x-intercepts if they exist.

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