Which of the following equations has two dist | vedclass

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(C) For a quadratic equation $ax^{2}+bx+c=0$,the roots are distinct and real if the discriminant $D = b^{2}-4ac > 0$.
$(a)$ For $2x^{2}-3\sqrt{2}x

Vedclass · Accepted Answer

$x^{2}+x-5=0$

Vedclass · Answer

(C) For a quadratic equation $ax^{2}+bx+c=0$,the roots are distinct and real if the discriminant $D = b^{2}-4ac > 0$. $(a)$ For $2x^{2}-3\sqrt{2}x+\frac{9}{4}=0$,$a=2, b=-3\sqrt{2}, c=\frac{9}{4}$. $D = (-3\sqrt{2})^{2}-4(2)(\frac{9}{4}) = 18-18 = 0$. Since $D=0$,the roots are real and equal. $(b)$ For $x^{2}+3x+2\sqrt{2}=0$,$a=1, b=3, c=2\sqrt{2}$. $D = (3)^{2}-4(1)(2\sqrt{2}) = 9-8\sqrt{2} < 0$. Since $D < 0$,the roots are not real. $(c)$ For $x^{2}+x-5=0$,$a=1, b=1, c=-5$. $D = (1)^{2}-4(1)(-5) = 1+20 = 21$. Since $D>0$,the roots are distinct and real. $(d)$ For $5x^{2}-3x+1=0$,$a=5, b=-3, c=1$. $D = (-3)^{2}-4(5)(1) = 9-20 = -11 < 0$. Since $D < 0$,the roots are not real. Thus,the correct option is $(c)$.

Which of the following equations has two distinct real roots?

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