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I was reading slides about the cancellation error in quadratic equations and it's written:

The roots of the quadratic equation:

$$x^2 - 2bx + c = 0$$

with $b^2 > c$ are given by $b \pm \sqrt{b^2 - c}$.

Fact that let me perplexed, since I always thought that the roots can be found using the following formula:

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

What the relation between one and the other?

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    $\begingroup$ So replace $b$ by $-2b$ and $a$ by 1 in the usual formula and what do you get? $\endgroup$ – almagest Apr 10 '16 at 13:57
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    $\begingroup$ Set $a=1$ and replace $b$ by $2b$ in the standard formula and do some algebra ... $\endgroup$ – Ethan Bolker Apr 10 '16 at 13:58
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If we write the quadratic like this $$ax^2+bx+c=0,$$ then the quadratic formula is right as you have written it; the roots are: $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ But the quadratic $$x^2-2bx+c=0$$ has replaced $b$ with $-2b$ and $a$ with $1$. This makes the quadratic $$\frac{2b\pm\sqrt{(-2b)^2-4(1)c}}{2(1)}=\frac{2b\pm2\sqrt{b^2-c}}{2}=b\pm\sqrt{b^2-c}.$$

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One formula solves the equation $x^2-2bx+c=0$, the other solves $ax^2+bx+c=0$.

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Use the usual formula for your equation $\;x^2-2bx+c=0\;$ :

$$x_{1,2}=\frac{2b\pm\sqrt{4b^2-4c}}2=\frac{2b\pm2\sqrt{b^2-c}}2=b\pm\sqrt{b^2-c}$$

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They use the so-called reduced formulae, when the coefficient of $x$ has the form $b=2b'$. In this case the discriminant is $\;\Delta=4b'^2-4ac=4\Delta'\;$ (reduced discriminat), and the formulae become $$x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-2b'\pm2\sqrt{\Delta'}}{2a}=\color{red}{\frac{-b'\pm\sqrt{\Delta'}}{a}}$$ Add the fact that here we have a monic polynomial ($a=1$) and you gete the mentioned formula.

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With the usual formula

$$x^2-2bx+c=0$$ has the roots

$$\frac{-(-2b)\pm\sqrt{(-2b)^2-4\cdot1\cdot c}}{2\cdot1}=b\pm\sqrt{b^2-c}.$$

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  • $\begingroup$ Crossed with @nbro. $\endgroup$ – Yves Daoust Apr 10 '16 at 14:11
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Never mind, they can be found by that formula only for this quadratic equation $$x^2 - 2bx + c = 0$$

Indeed if we use the usual formula:

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

to the equation above we obtain:

$$\frac{-(-2b) \pm \sqrt{(-2b)^2 - 4*1*c}}{2*1}$$

$$\frac{2b \pm \sqrt{4b^2 - 4c}}{2}$$

$$\frac{2b \pm \sqrt{4(b^2 - c)}}{2}$$

$$\frac{2b \pm 2\sqrt{b^2 - c}}{2}$$

$$\frac{2(b \pm \sqrt{b^2 - c})}{2}$$

$$b \pm \sqrt{b^2 - c}$$

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