The formula for solving a quadratic equation \(ax^2+bx+c=0\) is given as:
\(x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)
\(\text{Sum of roots}=\alpha+\beta=-\dfrac{b}{a}\)
\(\text{Product of roots}=\alpha\beta=\dfrac{c}{a}\)
The quadratic equation with roots \(\alpha\) and \(\beta\) can be written as:
\(x^2-(\text{sum of roots})x+(\text{product of roots})=0\)
Solve the equation \(x^2+4x-7=0\) by using completing the square method.
Based on equation \(x^2+4x-7=0\),
\(a=1\), \(b=4\), \(c=-7\).
Move the constant term, \(c\) to the right hand side of the equation,
\(\begin{aligned} x^2+4x-7&=0 \\ x^2+4x&=7. \end{aligned}\)
Add the term \(\left( \dfrac{b}{2} \right)^2\) on the left and right hand side of the equation,
\(\begin{aligned} x^2+4x+\left( \dfrac{4}{2} \right)^2&=7+\left( \dfrac{4}{2} \right)^2 \\ x^2+4x+2^2&=7+2^2\\ (x+2)^2&=11\\ x+2&=\pm \sqrt{11}. \end{aligned}\)
\(x=-5.317\) or \(x=1.317\).
Hence, the solutions of the equation \(x^2+4x-7=0\) are \(-5.317\) and \(1.317\).
Solve the equation \(2x^2-2x-3=0\) by using formula.
Based on the equation \(2x^2-2x-3=0\),
\(a=2\), \(b=-2\), \(c=-3\).
Use the formula for solving quadratic equation,
\(\begin{aligned} x&=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(2)(-3)}}{2(2)} \\ &=\dfrac{2\pm \sqrt{28}}{4} \end{aligned}\)
\(\begin{aligned} x&=\dfrac{2-\sqrt{28}}{4} \\ &=-0.823 \end{aligned}\) or \(\begin{aligned} x&=\dfrac{2+\sqrt{28}}{4} \\ &=1.823 .\end{aligned}\)
Hence, the solutions of the equation \(2x^2-2x-3=0\) are \(-0.823\) and \(1.823\).
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