Question

I really need help with solving quadratics using the quadratic formula, if anybody could help me with these problems that would be awesome, i would like an explanation on how to do this.

Answer 1

Answer:

1. Solutions: x = 3.4, 10.6

2. Solutions: $\textsf{ x = \dfrac{-7}{2} } + \dfrac{i\sqrt{15}}{2}$, $\textsf{ x = \dfrac{-7}{2} } - \dfrac{i\sqrt{15}}{2}$

3. Solutions: x = -1, 11

Step-by-step explanation:

Quadratic formula:  $\large \textsf{ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} }$

Quadratic equation: ax² + bx c = 0, where a ≠ 0

1. x² - 14x + 36 = 0

a = 1, b = -14, c = 36

Substitute the given values into the formula:

$\normalsize \textsf{ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} }\\\\\normalsize \textsf{ x = \dfrac{-(-14) \pm \sqrt{(-14)^2-4(1)(36)}}{2(1)} }\\\\\normalsize \textsf{ x = \dfrac{14 \pm \sqrt{196-144}}{2} }\\\\\normalsize \textsf{ x = \dfrac{14 \pm \sqrt{52}}{2} }\\\\\normalsize \textsf{ x = \dfrac{14 \pm 7.21}{2} }$

Separate into two cases:

$\normalsize \textsf{ x = \dfrac{14 + 7.21}{2} }\\\\\normalsize \textsf{ x = \dfrac{21.21}{2} }\\\\\normalsize \textsf{ x = 10.605 }\\\\\textsf{Nearest tenth: 10.6}$          and          $\normalsize \textsf{ x = \dfrac{14 - 7.21}{2} }\\\\\normalsize \textsf{ x = \dfrac{6.79}{2} }\\\\\normalsize \textsf{ x = 3.395 }\\\\\textsf{Nearest tenth: 3.4}$

2. x² + 7x + 16 = 0

a = 1, b = 7, c = 16

Substitute the given values into the formula:

$\normalsize \textsf{ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} }\\\\\normalsize \textsf{ x = \dfrac{-7 \pm \sqrt{7^2-4(1)(16)}}{2(1)} }\\\\\normalsize \textsf{ x = \dfrac{-7 \pm \sqrt{49-64}}{2} }\\\\\normalsize \textsf{ x = \dfrac{-7 \pm \sqrt{-15}}{2} }\\\\\normalsize \textsf{ x = \dfrac{-7 \pm i\sqrt{15}}{2} }$

Imaginary number rule: For any positive real number "k", $\large \textsf{ \sqrt{-k} = i\sqrt{k} }$

Notes: Two imaginary solutions indicate that the graph will not intersect the x-axis. As a result, it has no real roots.

Separate into two cases:

• $\normalsize \textsf{ x = \dfrac{-7 + i\sqrt{15}}{2} }\implies\normalsize \textsf{ x = \dfrac{-7}{2} } + \dfrac{i\sqrt{15}}{2}$

• $\normalsize \textsf{ x = \dfrac{-7 - i\sqrt{15}}{2} }\implies\normalsize \textsf{ x = \dfrac{-7}{2} } - \dfrac{i\sqrt{15}}{2}$

3. x² - 10x - 11 = 0

a = 1, b = -10, c = -11

Substitute the given values into the formula:

$\normalsize \textsf{ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} }\\\\\normalsize \textsf{ x = \dfrac{-(-10) \pm \sqrt{(-10)^2-4(1)(-11)}}{2(1)} }\\\\\normalsize \textsf{ x = \dfrac{10 \pm \sqrt{100+44}}{2} }\\\\\normalsize \textsf{ x = \dfrac{10 \pm \sqrt{144}}{2} }\\\\\normalsize \textsf{ x = \dfrac{10 \pm 12}{2} }$

Separate into two cases:

$\normalsize \textsf{ x = \dfrac{10 + 12}{2} }\\\\\normalsize \textsf{ x = \dfrac{22}{2} }\\\\\normalsize \textsf{ x = 11 }$          and          $\normalsize \textsf{ x = \dfrac{10 - 12}{2} }\\\\\normalsize \textsf{ x = \dfrac{-2}{2} }\\\\\normalsize \textsf{ x = -1 }$

Hope this helps!

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