Solving Quadratic Equations with the Quadratic Formula
Answer:
[tex]x = 3[/tex] and [tex]x = \frac{1}{2}\\[/tex]
Step-by-step Explanation:
Recall:
if we have a quadratic equation, [tex]ax^2 +bx +c = 0[/tex], where [tex]a[/tex], [tex]b[/tex] and [tex]c[/tex] are real numbers and [tex]a \neq 0[/tex], [tex]x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex].
Given:
[tex]2x^2 -7x +3 = 0[/tex]
Solving for [tex]x[/tex]:
[tex]x = \frac{-(-7) \pm \sqrt{(-7)^2 -4(2)(3)}}{2(2)} \\ x = \frac{-(-7) \pm \sqrt{49 -4(2)(3)}}{2(2)} \\ x = \frac{7 \pm \sqrt{49 -24}}{4} \\ x = \frac{7 \pm \sqrt{25}}{4} \\ x = \frac{7 \pm 5}{4}[/tex]
Solving with the positive value:
[tex]x = \frac{7 +5}{4} \\ x = \frac{12}{4} \\ x = 3 [/tex]
Solving with the negative value:
[tex]x = \frac{7 -5}{4} \\ x = \frac{2}{4} \\ x = \frac{1}{2}[/tex]
