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Show that the equation `2x^(2)-6x+3=0` has real roots and find these roots. |
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Answer» The given equation is `2x^(2)-6x+3=0.` This is of the form `ax^(2)+bx+c=0,` where a=2, b=-6 and c=3. `:." "D=(b^(2)-4ac)={(-6)^(2)-4xx2xx3}=(36-24)=12gt0.` So, the given equation has real unequal roots. Solving `2x^(2)-6x+3=0` by quadratic formula, we have `x=(6+-sqrt(36-4xx2xx3))/((2xx2))=(6+-sqrt(36-24))/(4)=(6+-sqrt(12))/(4)` `implies" "x=(6+-2sqrt(3))/(4)impliesx=(3+-sqrt(3))/(2).` So, `((3+sqrt(3)))/(2)" and "((3-sqrt(3)))/(2)` are the roots of the given equation. |
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