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$$X^2-49=0$$

$$X^2=49$$

$$\sqrt{X}= ± \sqrt{49}$$

$$X=±7$$

Discriminant: $$B^2-4ac$$

$$A=1$$

$$B=0$$

$$C=49$$

$$(0)^2-4(1)(49)$$ $$0-196=(-196)$$

A negative discriminant means that the roots of the equation is not real, but as I have shown, the solution is clearly $±7$ which are both real and equal, and would mean that the discriminant should be equal to 0 as well.

Please help me. And God Bless

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  • $\begingroup$ When you string all your equations together with no separators it becomes unreadable. In any case, if your quadratic is $x^2-49$ then the discriminant is $0^2-4\times 1 \times (-49)=4\times 49>0$. $\endgroup$
    – lulu
    Commented Oct 15, 2020 at 13:25
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    $\begingroup$ $c=-49$, not $49$ $\endgroup$
    – Lion Heart
    Commented Oct 15, 2020 at 13:26
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    $\begingroup$ "which are both real and equal": in my country, $-7$ and $+7$ are distinct. $\endgroup$
    – user65203
    Commented Oct 15, 2020 at 13:58
  • $\begingroup$ Yves Daoust, yes you are right, and thank you for your insight with the problem I presented. I simply forgot to put "in their absolute values", but nonetheless, thank you very much. $\endgroup$
    – Link Ed
    Commented Oct 15, 2020 at 23:31

3 Answers 3

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Note that $c = -49$, so $\Delta = b^2 - 4ac = 0^2 + 4(1)(49) = \boxed{196}$

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The equation is

$$1\cdot x^2+0\cdot x+(-49)=0.$$

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If the quadratic equation is given by $$ax^2+bx+c=0$$ then, $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ The discriminant is $D=\sqrt{b^2-4ac}$

The equation in your question can be re-written as: $$x^2+0.x+(-49)=0$$

On comparing the coefficients of the terms, you'll get where you went wrong while finding the determinant.

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  • $\begingroup$ Thank You for all of you, It was really late here yesternight thus I couldn't keep focused anymore, and simply decided to ask much knowledgeable persons(all of you) for help. It is fully appreciated your efforts and prudence. Again Thank You and God Bless. $\endgroup$
    – Link Ed
    Commented Oct 15, 2020 at 23:26

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