In Delta ABC,m angle A = 90^{circ}. If AB = 3 | vedclass

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(B) Since $\Delta ABC$ is a right-angled triangle with $m \angle A = 90^{\circ}$,by the Pythagorean theorem,we have:$AB^2 + AC^2 = BC^2$Substituting t

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$4$

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(B) Since $\Delta ABC$ is a right-angled triangle with $m \angle A = 90^{\circ}$,by the Pythagorean theorem,we have:$AB^2 + AC^2 = BC^2$Substituting the given expressions:$(3x - 2)^2 + (5x + 4)^2 = (6x + 2)^2$Expanding the squares:$(9x^2 - 12x + 4) + (25x^2 + 40x + 16) = (36x^2 + 24x + 4)$Combining like terms on the left side:$34x^2 + 28x + 20 = 36x^2 + 24x + 4$Rearranging the terms to one side:$36x^2 - 34x^2 + 24x - 28x + 4 - 20 = 0$$2x^2 - 4x - 16 = 0$Dividing the entire equation by $2$:$x^2 - 2x - 8 = 0$Factoring the quadratic equation:$(x - 4)(x + 2) = 0$This gives $x = 4$ or $x = -2$.Since lengths must be positive,$3x - 2 > 0$ implies $x > 2/3$. Thus,$x = 4$ is the only valid solution.

In $\Delta ABC$,$m \angle A = 90^{\circ}$. If $AB = 3x - 2$,$AC = 5x + 4$,and $BC = 6x + 2$,find the value of $x$.

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