Step-by-step explanation:
For a quadratic in the form of [tex]ax^{2} + bx + c = 0[/tex], the solutions for [tex]x[/tex] can be found using the quadratic formula:
[tex]\frac{-b ± \sqrt{b^{2} - 4ac}}{2a}[/tex]
(1) [tex]m^{2} - 5m - 14 = 0[/tex]
First, let's get the values for [tex]a, b, c[/tex]:
[tex]a = 1[/tex]
[tex]b = -5[/tex]
[tex]c = -14[/tex]
Next, let's plug them into the quadratic equation:
[tex]\frac{-(-5) ± \sqrt{(-5)^{2} - 4(1)(-14)}}{2(1)}[/tex]
[tex]\frac{5 ± \sqrt{25 + 56}}{2}[/tex]
[tex]\frac{5 ± \sqrt{81}}{2}[/tex]
[tex]\frac{5 ± 9}{2}[/tex]
With this, we know the two solutions for [tex]m[/tex] are
[tex]\frac{5 + 9}{2} = \frac{14}{2} = 7[/tex]
[tex]\frac{5 - 9}{2} = \frac{-4}{2} = -2[/tex]
(2) [tex]b^{2} - 4b + 4 = 0[/tex]
First, let's get the values for [tex]a, b, c[/tex]:
[tex]a = 1[/tex]
[tex]b = -4[/tex]
[tex]c = 4[/tex]
Next, let's plug them into the quadratic equation:
[tex]\frac{-(-4) ± \sqrt{(-4)^{2} - 4(1)(4)}}{2(1)}[/tex]
[tex]\frac{4 ± \sqrt{16 - 16}}{2}[/tex]
[tex]\frac{4 ± \sqrt{0}}{2}[/tex]
[tex]\frac{4}{2}[/tex]
With this, we know the only solution for [tex]b[/tex] is [tex]2[/tex].
