Answer :
Answer:
[tex]x = 1 \pm i\, \sqrt{6}[/tex].
Step-by-step explanation:
Given a quadratic equation [tex]a\, x^{2} + b\, x + c = 0[/tex] where [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex] are constants ([tex]a \ne 0[/tex],) the quadratic formula gives the value of the two roots of this equation:
[tex]\displaystyle x = \frac{-b \pm \sqrt{b^{2} - 4\, a\, c}}{2\, a}[/tex].
The expression [tex](b^{2} - 4\, a\, c)[/tex] is the quadratic determinant of this quadratic equation. Depending on the sign of this determinant, the equation would have:
- Two real roots, if the determinant is positive,
- One real root, if the determinant is [tex]0[/tex], or
- Two complex roots, if the determinant is negative.
For the equation this question, [tex]a = 1[/tex], [tex]b = (-2)[/tex], and [tex]c = 7[/tex]. Quadratic determinant would be [tex](-24)[/tex] for this equation. Since the determinant is negative, the two roots of this equation would be complex numbers.
Specifically, note that [tex]\sqrt{-24} = \left(\sqrt{-1}\right)\, \left(\sqrt{24}\right) = 2\, i\, \sqrt{6}[/tex], where [tex]i[/tex] is the imaginary unit. Hence, the two roots of this equation would be:
[tex]\begin{aligned} x &= \frac{-b \pm \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{-(-2) \pm \left(2\, i\, \sqrt{6}\right)}{2} \\ &= 1 \pm i\, \sqrt{6}\end{aligned}[/tex].
Answer:
[tex]x=1 \pm i\sqrt{6}[/tex]
Step-by-step explanation:
The quadratic formula is an equation used to find the solutions of a quadratic equation in the form ax² + bx + c = 0:
[tex]\boxed{\begin{array}{l}\underline{\sf Quadratic\;Formula}\\\\x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\textsf{when} \;ax^2+bx+c=0 \\\end{array}}[/tex]
In the case of x²- 2x + 7 = 0, the values of a, b and c are:
- a = 1
- b = -2
- c = 7
Substitute these values into the quadratic formula:
[tex]x=\dfrac{-(-2) \pm \sqrt{(-2)^2-4(1)(7)}}{2(1)}\\\\\\\\x=\dfrac{2 \pm \sqrt{4-28}}{2}\\\\\\\\x=\dfrac{2 \pm \sqrt{-24}}{2}\\\\\\\\[/tex]
Rewrite -24 as the product of 2², -1 and 6:
[tex]x=\dfrac{2 \pm \sqrt{2^2\cdot-1\cdot 6}}{2}[/tex]
[tex]\textsf{Apply the radical rule:} \quad \sqrt{ab}=\sqrt{\vphantom{b}a}\sqrt{b}[/tex]
[tex]x=\dfrac{2 \pm \sqrt{2^2}\sqrt{-1}\sqrt{6}}{2}[/tex]
[tex]\textsf{Apply the radical rule:} \quad \sqrt{a^2}=a, \quad a \geq 0[/tex]
[tex]x=\dfrac{2 \pm 2\sqrt{-1}\2\sqrt{6}}{2}[/tex]
Cancel the common factor 2:
[tex]x=1 \pm \sqrt{-1}\2\sqrt{6}[/tex]
[tex]\textsf{Apply the imaginary number rule:} \quad \sqrt{-1}=i[/tex]
[tex]x=1 \pm i\sqrt{6}[/tex]
Therefore, the solutions to x²- 2x + 7 = 0 are:
[tex]\Large\boxed{\boxed{x=1 \pm i\sqrt{6}}}[/tex]
