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I tried using the quadratic formula but I don't have a $c$. Do I just put a $1$ in its place or something? The answer is supposed to be $\mp \frac15 \sqrt{15}$.

Rócherz
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user591195
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    Well, you have a $c$...just not a $b$. The standard form of a quadratic is $ax^2+bx+c$ and here $b=0$. But the quadratic formula is overkill here...your equation is just $x^2=\frac 35$. – lulu Oct 28 '18 at 13:29
  • thanks, my teacher skipped this one and I had no idea how to do it – user591195 Oct 28 '18 at 13:46

4 Answers4

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$$5x^2-3=0 \implies x^2=3/5 \implies x=\pm \sqrt{3/5}=\pm{1 \over 5}\sqrt{15}$$

MathIsNice1729
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$$\frac{-0\pm\sqrt{0^2+4\cdot5\cdot3}}{2\cdot 5}.$$

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Method $1$:

$a=5$ clearly, while you have $c=-3$. There's no $x$ term, so $b=0$, and thus we have: $$x=\frac{0\pm\sqrt{0^2-4\cdot5\cdot(-3)}}{2\cdot5}=\pm\frac{\sqrt{60}}{10}=...$$

Method $2$: This is difference of two squares: $(a^2-b^2)=(a+b)(a-b)$, so we have: $$(5x^2-3)=(x\sqrt5+\sqrt3)(x\sqrt5-\sqrt3)=0$$ $$\to x=\pm\frac{\sqrt3}{\sqrt5}$$

You can simplify both of these to show they are the same.

Rhys Hughes
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An option:

$x^2-3/5=0;$

$(x-\sqrt{3/5})(x+ \sqrt{3/5})=0$;

Note: For the above product to be zero,

one factor must be zero.

$x_1=\sqrt{3/5}$; $x_2=-\sqrt{3/5}.$

Peter Szilas
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