(C) Comparing $x^2-x+1$ with the general quadratic form $ax^2+bx+c$,we have $a=1$,$b=-1$,and $c=1$.The discriminant $D$ is calculated as $D = b^2 - 4a

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(C) Comparing $x^2-x+1$ with the general quadratic form $ax^2+bx+c$,we have $a=1$,$b=-1$,and $c=1$.The discriminant $D$ is calculated as $D = b^2 - 4ac$.Substituting the values,$D = (-1)^2 - 4(1)(1) = 1 - 4 = -3$.Since the discriminant $D < 0$,the quadratic expression has no real roots and therefore has no proper linear factors over the set of real numbers.

Competitive Quantitative Aptitude Algebra QUADRATIC EQUATION

Medium

The expression $x^2-x+1$ has:

A
one proper linear factor
B
two proper linear factors
C
no proper linear factor
D
None of these

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QUADRATIC EQUATION

If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 2x + 4 = 0$,then $\frac{1}{\alpha^3} + \frac{1}{\beta^3}$ is equal to

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If the equations $x^{2}+2x-3=0$ and $x^{2}+3x-k=0$ have a common root,then the non-zero value of $k$ is:

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The least positive value of $a$ for which the equation $2x^{2} + (a - 10)x + \frac{33}{2} = 2a$ has real roots is

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Solve the given two equations and provide the correct answer from the given options.
$I.$ $x^{2}+14x+49=0$
$II.$ $y^{2}+9y=0$

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If $\alpha, \beta$ are the roots of the equation $x^{2}+kx+12=0$ such that $\alpha-\beta=1$,the value of $k$ is

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The expression $x^2-x+1$ has:

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If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 2x + 4 = 0$,then $\frac{1}{\alpha^3} + \frac{1}{\beta^3}$ is equal to

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Solve the given two equations and provide the correct answer from the given options.
$I.$ $x^{2}+14x+49=0$
$II.$ $y^{2}+9y=0$