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Quadratic Equation

Unit: 4
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Class 10: Optional Math

Quadratic Equation, Conversion of Quadratic Equation form

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    Quadratic Equation

    The equation $ax^2+ bx + c = 0$ (a ≠ 0) is called a quadratic equation in one variable. The graph of the equation ax2 + bx + c = 0 is in the shape of ∪ or ∩ , which is called a parabola. The shape of the graph of a quadratic equation depends on the values of the real numbers a, b, and c. The point where the direction of the parabola changes is called the turning point. The turning point of the parabola is also called the vertex of the parabola.

     

    Conversion of Quadratic Equation form

    Conversion of Quadratic Equation $y = ax^2 + bx + c$ into $y = a(x - h)^2 + k$

    To convert the quadratic function $y = ax^2 + bx + c$ in the form of $y = a(x - h)^2 + k$, the following method of completing the square is used:

    $ \begin{aligned} y &= ax^2 + bx + c \\[6pt] \text{or, } y &= a\left\{x^2 + \frac{bx}{a} + \frac{c}{a}\right\} \\[6pt] \text{or, } y &= a\left\{x^2 + 2 \cdot x \cdot \frac{b}{2a} + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a}\right\} \\[6pt] \text{or, } y &= a\left\{\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a}\right\} \\[6pt] \text{or, } y &= a\left\{\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b^2 - 4ac}{4a^2}\right)\right\} \\[6pt] \text{or, } y &= a\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b^2 - 4ac}{4a}\right) \\[6pt] \text{or, } y &= a\left\{x - \left(-\frac{b}{2a}\right)\right\}^2 + \left(\frac{4ac - b^2}{4a}\right) \quad \ldots \text{(i)} \end{aligned} $

    Now, comparing equation (i) with: $y = a(x - h)^2 + k \quad \ldots \text{(ii)}$

    We get: $h = -\frac{b}{2a}, \quad k = \frac{4ac - b^2}{4a}$

    Thus, the vertex of the parabola is: $$\boxed{(h, k) = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right)}$$

    And the axis of the parabola is: $$\boxed{x = -\frac{b}{2a}}$$

    Key Formulas Summary

    ComponentFormula
    Vertex$\left(-\dfrac{b}{2a}, \dfrac{4ac - b^2}{4a}\right)$
    Axis of symmetry$x = -\dfrac{b}{2a}$
    h$-\dfrac{b}{2a}$
    k$\dfrac{4ac - b^2}{4a}$

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