Question

quadratic equationsplease can anyone answer this please please please

Answer 1

Let the altitude be x,
base = 4+x

$area = \frac{1}{2} \times b \times h$
48= 1/2 × (4+x) × x
48 = (4+x) × x/2
96 = 4x + x²
x²+4x-96 = 0
x²+12x-8x - 96 = 0
x(x+12) - 8 (x+12) = 0
(x+12) (x-8) = 0

x+12 = 0
x= -12

x-8 = 0
x = 8

base = 4+x ; altitude = 8cm
= 4 +8
base = 12cm

Answer 2

let base = b and altitude= h.

now, b= h+4



area of triangle = 1/2 × b×h
48 = 1/2 x (h+4) x h
96 = h^2+ 4h
0 = h^2 + 4h -96
0 = h^2+ 12h - 8h -96
0 = h(h+12) - 8(h+12)
0 = (h+12) (h-8)

so either h+12 = 0 or h-8 =0

so h= -12 cm or h= 8 cm.
as length cannot be negative
so h= 8 cm and b=h+4 = 12cm


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