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I'm helping my child with his homework. One of the problems is this: Determine the solution set of the equation

$x^2+3x+7=4$

Here is my attempt to determine $x_1$ and $x_2$:

$a^2+2ab+b^2=(a+b)^2$

$2ab=3x$

$2xb=3x$

$2b=3$

$b=\frac{3}{2}=1.5$

$b^2=2.25$

$x^2+3x+b^2-b^2+7=4$

$x^2+3x+2.25-2.25+7=4$

$(x^2+3x+2.25)-2.25+7=4$

$(x^2+3x+2.25)+7=4+2.25$

$(x^2+3x+2.25)+7=6.25$

$(x^2+3x+2.25)=6.25-7$

$(x^2+3x+2.25)=-0.75$

$(x+1.5)^2=-0.75$

In other similar equations we would take the square root of both sides. But I cannot take a square root of a negative number ($-0.75$).

How can I solve it, i. e. find out the values of $x$?

Glory to Russia
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  • There are no real solutions. Only complex ones. Same would be for $x^2+1=0$ – Andrei Mar 25 '20 at 22:09
  • If you did this right (didn't check; my eyes glazed over) then if you get a situation where you have to get a square root of a negative number then there is not solution. That happens often. You simply say there is nosolution. – fleablood Mar 25 '20 at 22:11
  • By the way, are you familiar with completing the square? It seems you went to a convoluted method of trying to re-invent it by scratch. $x^2 + 3x +7 =4\implies x^2 + 3x = -3\implies x^2 + 2*\frac 32x + (\frac 32)^2 = -3+(\frac 32)^2\implies(x+\frac 32)^2 = -\frac 34$ which is impossible. There is no solution. ... it happens.... – fleablood Mar 25 '20 at 22:15
  • Also a pet peeve of mine. Writing things as decimals are not considered any more "real" or correct and they are a heck of a lot harder. It'd be a lot easier both to do and to read this if you used fractions. $b=\frac 32$ and $b^2 = \frac 94$ and $(x^2 + 3x + \frac 94) = 4-7+\frac 94 = -3 +\frac 94 = -\frac 34$ and so on. – fleablood Mar 25 '20 at 22:23

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This is actually much simpler if you subtract the 4 to the left hand side first: $$x^2 + 3x + 3 = 0$$ Then the solution set must satisfy the quadratic formula: $$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ where we have the quadratic $ax^2 + bx + c = 0$. In this case, $$a=1, b=3, c = 3$$ So then $x$ must be $$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(3)}}{2a}$$ However, we notice here that inside the square root, we have a negative number $(-3)$, which indicates that the quadratic has no solution set in the real numbers.

If we are allowed to use imaginary numbers, then the solution would be $$x = \frac{-3\pm \sqrt{-3}}{2} = \frac{-3\pm i\sqrt{3}}{2}$$

BSplitter
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