Question

Hey can anyone pls answer these math problems!!

Answer 1

Problem 1

We have something in the form $ax^2+bx+c = 0$, where

• a = 1
• b = -8
• c = 15

Those values are plugged into the quadratic formula as shown below.

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-8)\pm\sqrt{(-8)^2-4(1)(15)}}{2(1)}\\\\x = \frac{8\pm\sqrt{4}}{2}\\\\x = \frac{8\pm2}{2}\\\\x = \frac{8+2}{2} \ \text{ or } \ x = \frac{8-2}{2}\\\\x = \frac{10}{2} \ \text{ or } \ x = \frac{6}{2}\\\\x = 5 \ \text{ or } \ x = 3$

The roots or solutions are x = 5 and x = 3

To check these answers, plug them back into the original equation and you should get the same number on both sides. You can also use a graphing calculator to confirm the answers. In this case, the parabola crosses the x axis at 5 and 3.

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Problem 2

This time we have

• a = 1
• b = 14
• c = 0

which leads to:

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(14)\pm\sqrt{(14)^2-4(1)(0)}}{2(1)}\\\\x = \frac{-14\pm\sqrt{196}}{2}\\\\x = \frac{-14\pm14}{2}\\\\x = \frac{-14+14}{2} \ \text{ or } \ x = \frac{-14-14}{2}\\\\x = \frac{0}{2} \ \text{ or } \ x = \frac{-28}{2}\\\\x = 0 \ \text{ or } \ x = -14$

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Problem 3

We'll plug in a = 1, b = -5, and c = 1

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-5)\pm\sqrt{(-5)^2-4(1)(1)}}{2(1)}\\\\x = \frac{5\pm\sqrt{21}}{2}\\\\x = \frac{5+\sqrt{21}}{2} \ \text{ or } \ x = \frac{5-\sqrt{21}}{2}$

Unlike the previous results, we cannot simplify the square roots down to some whole number. So we leave the roots as they are.

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Problem 4

We first need to move the 12 over to the left side so that we have 0 on the right side.

The equation $x^2-x = 12$ becomes $x^2-x-12 = 0$ at which point we have

• a = 1
• b = -1
• c = -12

So,

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-12)}}{2(1)}\\\\x = \frac{1\pm\sqrt{49}}{2}\\\\x = \frac{1\pm7}{2}\\\\x = \frac{1+7}{2} \ \text{ or } \ x = \frac{1-7}{2}\\\\x = \frac{8}{2} \ \text{ or } \ x = \frac{-6}{2}\\\\x = 4 \ \text{ or } \ x = -3$

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Problem 5

Start by subtracting 5 from each side

The equation $x^2-10x+14 = 5$ becomes $x^2-10x+9 = 0$

We'll plug in a = 1, b = -10, c = 9

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-10)\pm\sqrt{(-10)^2-4(1)(9)}}{2(1)}\\\\x = \frac{10\pm\sqrt{64}}{2}\\\\x = \frac{10\pm8}{2}\\\\x = \frac{10+8}{2} \ \text{ or } \ x = \frac{10-8}{2}\\\\x = \frac{18}{2} \ \text{ or } \ x = \frac{2}{2}\\\\x = 9 \ \text{ or } \ x = 1$

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Problem 6

This is similar to the first three problems where the equation is already in $ax^2+bx+c = 0$ form.

This time we have a = 1, b = 5, c = 3.

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(5)\pm\sqrt{(5)^2-4(1)(3)}}{2(1)}\\\\x = \frac{-5\pm\sqrt{13}}{2}\\\\x = \frac{-5+\sqrt{13}}{2} \ \text{ or } \ \frac{-5-\sqrt{13}}{2}$