How do I find the solution(s) to my second-degree equation?

Question

Solve the second-degree equation $x^2 - 3x=4$

My attempt : $$ \begin{align} x^2-3x &= 4\\ x(x-3) &= 4\\ x-3 &= 4 \\ x &= 7\\ \end{align} $$

I managed to solve one part of this problem but that one part is wrong. I don't understand how to solve for $x$, when $x(x-3) = 4$. I can solve for the $x$ inside the bracket but I don't know how to solve the one outside the bracket and the $x$ I solved for inside the bracket is somehow wrong.

Answer 1

$$x(x-3)=4$$ $$x^2-3x-4=0$$ $$(x-4)(x+1)=0$$

Are you able to complete the rest?

Extra comment: In your working, from $x(x-3)=4$, we cannot conclude that $(x-3)=4$.

Extra comment 2: note that $$(x-a)(x-b)=x^2-(a+b)x+ab$$

Hence to recover $a$ and $b$, I tried for factor of $-4$ where the sum is equal to $3$.

For general quadractic equation, you can complete the square or use the quadractic formula.

Answer 2

$x^2-3x=4$ can be solved by using the pq-formula :

$x^2-3x-4=0$

$x_1=\frac{3}{2}+\sqrt{\frac{9}{4}+4}=4$

$x_2=\frac{3}{2}-\sqrt{\frac{9}{4}+4}=-1$

Answer 3

Guessing the solution can be hard if you're not used to it. Here's something to help you understand how SiongthyeGoh found it.

Let $f$ be defined by :

$$ \forall x \in \mathbb{R}, f(x) = x^2-3x $$

We want to find every $x$ so that $f(x)=4$. Let's solve it !

$$ \begin{align} x^2-3x &= 4& x^2-3x ~~\text{ begins like }~~ x^2-3x+\frac94=\left(x-\frac32\right)^2\\ \left(x-\frac32\right)^2-\frac94 &= 4 &\text{factorizing with this remarkable identity}~~~~~~~~\\ \left(x-\frac32\right)^2 &= \frac{25}4 & 4+\frac94= \frac{16}4+\frac94 = \frac{25}4~~~~~~~~~~~~~~~~~~~~~~~~\\ x-\frac32 &= \pm\sqrt{\frac{25}4} &\text{be careful with } \pm \text{ when removing the square !}\\ x &= \frac32 \pm\frac52 & \sqrt{25}=5 ~~\text{ and }~~ \sqrt{4}=2 ~~\text{ so }~~ \sqrt{\frac{25}4}=\frac52~~~~~~~~~\\ x = 4 ~~&\text{ or }~~ x = -1 & \frac32+\frac52=\frac82=4 ~~~~\text{ and }~~~~ \frac32-\frac52=\frac{-2}2=-1 \end{align} $$

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This is the best way that I know to solve equations like this where you need no theory. All you need to remember is that you need to do :

$$ x^2+bx~~~~ \longrightarrow ~~~~\left(x+\frac{b}2\right)^2-\left(\frac{b}2\right)^2 $$

This is called using the canonical form.

Answer 4

Hi I have found something new!!

I am going to tell you a new formula:

Any Quadratic Polynomial $ax^2$ + $bx$ + $c$ = $0$ , whose roots are $\alpha$ and $\beta$, You just find

$$D = (am)^2$$

for some complex $m$. D is the discriminant given by $b^2 -4ac$.

Then put $m$ in this equation

$$\alpha = (\frac{-b}{a} + m)\frac{1}{2}$$

$$\beta = (\frac{-b}{a} - m)\frac{1}{2}$$

You will get your roots!

Let's Try:

$$x^2- 3x- 4$$

$a=1$

$b= -3$

$c=-4$

Now,

$$D = (am)^2$$

$D = (-3)^2 - 4(1)(-4)$

$D = 9 + 16$

$D =25$

$$25=[(1)(m)]^2$$

$$25 = m^2$$

$$m =5$$

It's easy now

$$\alpha = (\frac{3}{1} + 5)\frac{1}{2}$$

$$\beta = (\frac{3}{1} - 5)\frac{1}{2}$$

The roots would be:

$$\alpha = 4$$ $$\beta = -1$$

You need to remember only these three

$$D = (am)^2$$ $$\alpha = (\frac{-b}{a} + m)\frac{1}{2}$$ $$\beta = (\frac{-b}{a} - m)\frac{1}{2}$$

Note: I did not take $m=\pm 5$. I am putting that in $\alpha$ and $\beta$. This gives even the Complex roots also, you can try.

I derived this formula yesterday

!!