Solving $ x^2-8x+15=0$?

Question

$x^2-8x+15=0$ I always had problems with equations. I tried solving it but I got weird numbers .. I would really appreciate some help and tips how to solve this

Answer 1

Generally, quadratic equation of the form $\color{orange}{a}x^2 - \color{blue}{b}x + \color{green}{c} = 0$ where $\color{orange}{a} \neq 0$ can be solve by evaluating $$\color{brown}{\Delta} = \color{blue}{b}^2 - 4\color{orange}{a}\color{green}{c}$$

Then, depending on the sign of $\color{brown}{\Delta}$

$$ \begin{array}{c|c|c} \color{brown}{\Delta} > 0 & \color{brown}{\Delta} = 0& \color{brown}{\Delta} < 0 \\ \hline \textrm{two solutions} & \textrm{one solution} & \textrm{no real solutions}\\ \hline x = \frac{-\color{blue}{b} \pm \sqrt{\color{brown}{\Delta}}}{2\color{orange}{a}} & x = \frac{-\color{blue}{b}}{2\color{orange}{a}}& - \end{array} $$

Then for your particular case we have $$\color{orange}{1}x^2 - \color{blue}{8}x + \color{green}{15} = 0$$

$$\color{brown}{\Delta} = (\color{blue}{-8})^2 - 4 \times \color{orange}{1} \times \color{green}{15}$$

$$\color{brown}{\Delta} = 4$$

Since $\color{brown}{\Delta} > 0$ we get that

$$x = \frac{-(\color{blue}{-8}) \pm \sqrt{\color{brown}{4}}}{2 \times \color{orange}{1}}$$

$$x = \frac{8 \pm 2}{2}$$ $$x = 4 \pm 1$$ $$x = 5 \qquad \vee\ \qquad x = 3$$

Answer 2

Recall the quadratic formula $$x=\frac {-b \pm \sqrt{b^2-4ac}}{2a}$$

You are faced with an equation of the form $$ax^2+bx+c=0$$ where $$a=1,\,b=-8,\,c=15$$ From there you only have to plug the values into the equation to get both roots $$x=\frac {-(-8) \pm \sqrt{(-8)^2-4(1)(15)}}{2(1)}$$ $$x=\frac {8\pm \sqrt{64-60}}{2}$$ $$x_1=\frac {8-\sqrt{4}}{2}=3$$ $$x_2=\frac {8+\sqrt4}{2}=5$$

When applying the formula $x=\frac {-b \pm \sqrt{b^2-4ac}}{2a}$, it is important to note that $x$ is a real number if and only if the discriminant $b^2-4ac$ is greater than or equal to zero. When the discriminant is negative you get a complex root involving the imaginary number $i$. Namely, if $0>b^2-4ac$ then, $x = a+bi$ such that $x$ is a complex number and $i =\sqrt{-1}$

Hope this helps.

Answer 3

In general, we can solve quadratic equations $ax^2+bx+c=0$ with the quadratic formula:

$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

As a hint, we can write $$x^2-8x+15 = 1\cdot x^2 + (-8)x + 15. $$

Now apply the quadratic formula.

Answer 4

$$x^2-8x+15=0$$

Think what two numbers multiply to make $15$ and add to make $-8$.

I know $5$ times $3$ is $15$. But $5$ plus $3$ is $8$. This is not $-8$ but it is close, so I change both numbers to negative, that way $-5+-3=-8$ and $(-5)(-3)=15$.

So my two numbers are $\color{red}{-3}$ and $\color{red}{-5}$. Then I look at what:

$$(x \color{red}{-3})(x \color{red}{-5})$$

Is, sure enough this is $x^2-8x+15$. This can be seen by "Foil" or double distributive.

So to solve $$(x-3)(x-5)=0$$

We either have $x-3=0$ and or $x-5=0$. So $x=3$ or $x=5$.

Answer 5

If you're looking to avoid the Quadratic Formula and you're new to quadratic equations, it is possible that your questions can be solved using the following (basic) method. When you have:

$$ax^2 + bx + c = 0,$$

with $a = 1$, a very elementary method to check for solutions is to find two numbers that multiply to $a*c$ and add to $b$. Here, you have $a = 1$, $b = -8$, $c = 15$. So, we want to two numbers which multiply to:

$(a*c) = (1*15) = 15$, and add to $b = (-8)$

Two such numbers are (I believe you can verify) $-5$ and $-3$. You can now rewrite your equation as $$(x-5)(x-3) = 0$$

And, clearly, the two values of x which make this equal $0$ are $$x = 5, x = 3$$ since $5-5 = 0$, and $3-3 = 0$, and multiplying any value by $0$ gives $0$.

I hope this helps. You won't always be able to (easily) find values that multiply to $a*c$ and add to $b$, so in that case you'll most likely need the quadratic formula, as mentioned above.