Check whether the equation 6x^2 - 7x + 2 = 0 | vedclass

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Question

(A) Given equation: $6x^2 - 7x + 2 = 0$. Comparing with $ax^2 + bx + c = 0$, we get $a = 6, b = -7, c = 2$. Discriminant $D = b^2 - 4ac = (-7)^2 - 4(6

Vedclass · Accepted Answer

Roots are real and are $1/2, 2/3$

Vedclass · Answer

(A) Given equation: $6x^2 - 7x + 2 = 0$. Comparing with $ax^2 + bx + c = 0$, we get $a = 6, b = -7, c = 2$. Discriminant $D = b^2 - 4ac = (-7)^2 - 4(6)(2) = 49 - 48 = 1$. Since $D > 0$, the equation has two distinct real roots. To solve by completing the square, divide by $6$: $x^2 - (7/6)x + 1/3 = 0$. Rewrite as $x^2 - 2(7/12)x = -1/3$. Add $(7/12)^2$ to both sides: $x^2 - 2(7/12)x + (7/12)^2 = -1/3 + 49/144$. $(x - 7/12)^2 = (-48 + 49) / 144 = 1/144$. Taking square root: $x - 7/12 = \pm 1/12$. Case $1$: $x = 7/12 + 1/12 = 8/12 = 2/3$. Case $2$: $x = 7/12 - 1/12 = 6/12 = 1/2$. Thus, the roots are $1/2$ and $2/3$.

Check whether the equation $6x^2 - 7x + 2 = 0$ has real roots, and if it has, find them by the method of completing the squares.

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