Prove that if ${b}^{2}-4c<0$, then there is no such $x\in R$ such that ${x}^{2}+bx+c=0$

Alexandra Richardson
2022-07-18
Answered

Prove that if ${b}^{2}-4c<0$, then there is no such $x\in R$ such that ${x}^{2}+bx+c=0$

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edgarovhg

Answered 2022-07-19
Author has **12** answers

Step 1

Two complex roots of a polynomial of degree 2:

$a{x}^{2}+bx+c$

are given by the formula

${x}_{1,2}=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

Step 2

and cannot be real for $\mathrm{\Delta}={b}^{2}-4ac<0$. In you case $a=1$

Two complex roots of a polynomial of degree 2:

$a{x}^{2}+bx+c$

are given by the formula

${x}_{1,2}=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

Step 2

and cannot be real for $\mathrm{\Delta}={b}^{2}-4ac<0$. In you case $a=1$

przesypkai4

Answered 2022-07-20
Author has **1** answers

Step 1

Suppose there were such x. Then

$0={x}^{2}+bx+c=(x-\frac{b}{2}{)}^{2}-\frac{{b}^{2}}{4}+c$

Step 2

So $(x-\frac{b}{2}{)}^{2}=\frac{{b}^{2}}{4}-c=\frac{1}{4}({b}^{2}-4c)<0$

which is not possible for real numbers.

Suppose there were such x. Then

$0={x}^{2}+bx+c=(x-\frac{b}{2}{)}^{2}-\frac{{b}^{2}}{4}+c$

Step 2

So $(x-\frac{b}{2}{)}^{2}=\frac{{b}^{2}}{4}-c=\frac{1}{4}({b}^{2}-4c)<0$

which is not possible for real numbers.

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