In Delta ABC,mangle B = 90^{circ},AB = 2x + 3 | vedclass

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(A) In $\Delta ABC$,since $m\angle B = 90^{\circ}$,by the Pythagorean theorem:$AC^2 = AB^2 + BC^2$$(3x - 1)^2 = (2x + 3)^2 + (x + 2)^2$$9x^2 - 6x + 1

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) In $\Delta ABC$,since $m\angle B = 90^{\circ}$,by the Pythagorean theorem:$AC^2 = AB^2 + BC^2$$(3x - 1)^2 = (2x + 3)^2 + (x + 2)^2$$9x^2 - 6x + 1 = (4x^2 + 12x + 9) + (x^2 + 4x + 4)$$9x^2 - 6x + 1 = 5x^2 + 16x + 13$$4x^2 - 22x - 12 = 0$Dividing by $2$:$2x^2 - 11x - 6 = 0$$2x^2 - 12x + x - 6 = 0$$2x(x - 6) + 1(x - 6) = 0$$(x - 6)(2x + 1) = 0$Thus,$x = 6$ or $x = -\frac{1}{2}$.Since side lengths must be positive,$x = -\frac{1}{2}$ is rejected because it results in negative side lengths.Therefore,$x = 6$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$,$AB = 2x + 3$,$BC = x + 2$,and $AC = 3x - 1$. Find the value of $x$.

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