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I used Mathway to find x in the equation

$-1.9497137966666855 * x^2 + 0.0 * x - 1 = -0.6556211074636167 * x^2 + 0.0*x - 5$

it gave the answer $1.75811509$ and $-1.75811509$

but when I replace one of these values with $x$, for example if I plug $x = 1.75811509$ into $-1.9497137966666855 * x^2 + 0.0 * x - 1$ which is

$$-1.9497137966666855 * 1.75811509^2 + 0.0*1.75811509 - 1$$

it gives the value of $-7.02650426$ instead of $0$ (zero)

abacaba
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Jaisal Francis
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    "instead of 0 (zero)" $;-;$ Why zero? The right-hand side of the original equation is not $0$. – dxiv Jun 02 '22 at 03:54
  • @dxiv rhs is -0.6556211074636167 * x^(2) + 0.0*x = 5 – Jaisal Francis Jun 02 '22 at 03:57
  • It's $-5$ not $=5$, and if you substitute $,x=1.75811509,$ there, you get $\dots$ – dxiv Jun 02 '22 at 03:59
  • @dxiv but when I calculate -0.6556211074636167 * x^(2) + 0.0*x, x = 1.75811509 I am getting -2.02650430235 instead of 5 – Jaisal Francis Jun 02 '22 at 04:05
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    The RHS of the original equation is $,-0.6556211074636167 \cdot x^2 + 0.0 \cdot x, -5$. When $x=1.75811509$ it evaluates to $−7.0265\dots$ which matches the LHS within the precision you are working with. Keep in mind that the equation is of the form $\text{LHS} = \text{RHS}$, not $\text{LHS} = \text{RHS} = 0$. – dxiv Jun 02 '22 at 04:12
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    @dxiv oh my bad I thought a=b=0 my bad. Thank you for your time – Jaisal Francis Jun 02 '22 at 04:18

1 Answers1

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You have, $$\rm{LHS}=-1.9497137966666855(1.75811509)^2-1=\color{blue}{-7.026504}2603506961$$ and $$\rm{RHS}=-0.6556211074636167(1.75811509)^2-5=\color{blue}{-7.026504}302354686$$ which indeed verifies $\rm{LHS}=\rm{RHS}$ upto certain precision. The $\rm{LHS}$ or the $\rm{RHS}$ doesn't have to be zero necessarily.

Hope this helps. :)

ultralegend5385
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