The area of a triangle is #2x^2-11x+15#. If the triangle's base is #2x-5#, what is its height?

Recall;

Area of triangle # = 1/2 "base" xx "height"#

Area of triangle #= 2x^2 - 11x + 15#

base #=2x - 5#

Plugging in the given;

#2x^2 - 11x + 15 = 1/2 (2x - 5) xx "height"#

#2x^2 - 11x + 15 = (2x - 5)/2 xx "height"#

#(2x^2 - 11x + 15)/1 = (2x - 5)/2 xx "height"#

#2(2x^2 - 11x + 15) = (2x - 5) xx "height"#

#4x^2 - 22x + 30 = (2x - 5) xx "height"#

#(4x^2 - 22x + 30)/(2x - 5) = "height"#

Resolving the quadratic equation;

#(4x^2 - 22x + 30)#

Simplifying;

#(4x^2)/2 - (22x)/2 + 30/2#

#2x^2 - 11x + 15#

Using Factorization Method..

#6 and 5# are factors..

#2x^2 - 6x - 5x + 15#

Grouping;

#(2x^2 - 6x) (- 5x + 15)#

#2x(x - 3) - 5(x - 3)#

#(x - 3) (2x - 5)#

Therefore;

#(4x^2 - 22x + 30)/(2x - 5) = "height"#

#color(white)(xxxxx)darr#

#(color(red)(2)(x - 3) (2x - 5))/(2x - 5) = "height"#

#(color(red)(2)(x - 3) cancel(2x - 5))/cancel(2x - 5) = "height"#

#color(red)(2)(x - 3) = "height"#

#"Area of triangle = (1/2) * (base * height)"#

#A_t = (1/2) b h#

#"Given : " A_t = 2x^2 - 11x + 15, b = (2x - 5)#

#h = ((2 * A_t) / b)#

#h = (2 * (2x^2 - 11x + 15)) / (2x - 5)#

#h = (2 * cancel(2x - 5) * (x - 3)) / cancel(2x - 5)#

#h = 2 * (x - 3)#