The roots of a quadratic equation ........... | vedclass

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Question

(B) quadratic equation has recurring (equal) roots if its discriminant $D = b^{2}-4ac$ is equal to $0$.For option $A$: $x^{2}-6x-8=0$,$D = (-6)^{2}-4(

Vedclass · Accepted Answer

$x^{2}-6 x+9=0$

Vedclass · Answer

(B) quadratic equation has recurring (equal) roots if its discriminant $D = b^{2}-4ac$ is equal to $0$.For option $A$: $x^{2}-6x-8=0$,$D = (-6)^{2}-4(1)(-8) = 36+32 = 68 
eq 0$.For option $B$: $x^{2}-6x+9=0$,$D = (-6)^{2}-4(1)(9) = 36-36 = 0$. Since $D=0$,the roots are recurring (equal).For option $C$: $x^{2}-9=0$,$D = (0)^{2}-4(1)(-9) = 36 
eq 0$.For option $D$: $x^{2}-4=0$,$D = (0)^{2}-4(1)(-4) = 16 
eq 0$.Therefore,the equation with recurring roots is $x^{2}-6x+9=0$.

The roots of a quadratic equation ............ are recurring.

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