Question

1: (7x-2)square


2: (5x-2)(3-x)

Answer 1

Step-by-step explanation

:1) (7x-2)^2

we know that (a-b)^2=a^2+b^2-2ab

(7x-2)^2=(7x)^2+2^2-2(7x)(2) =49x^2+4-28x

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

=5x(3-x)-2(3-x)

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

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

hope it's clearpls Mark me as brainliest

Answer 2

AnsWer:-

1.)

→(7x-2)²=(a-b)²

→(a-b)²=a²-2ab+b²

→(7x)²-2×7x×2+(2)²

→49x²28x+4=0

★Using Discriminant Formula★

๛a=49,b=-28,c=4

→d=b²-4ac

•Putting The Values•

→d=(-28)²-4×49×4

→d=700-784

→d=-84

[•°• Since d<0,there are no roots for the Equation]

$\rule{200}{1}$

2.)

→(5x-2)(3-x)

=>15x-5x²-6+2x

→5x²+17x-6=0

★Using Discriminant Formula★

๛a=5,b=17,c=-6

→d=b²-4ac

→d=(17)²-4×5×-6

→d=289+120

→d=409

[•°• Since d>0, there are roots for the Equation]

→√d=±√409

$\rule{200}{1}$

→Taking x(+)=$\frac{-b+ \sqrt{d}}{2a}$

→x=$\frac{-(17)+ \sqrt{409}}{2×5}$

→x=$\frac{-17+ \sqrt{409}}{10}$

$\rule{200}{1}$

→Taking x(-)=$\frac{-b- \sqrt{d}}{2a}$

→x=$\frac{-(17)- \sqrt{409}}{2×5}$

→x=$\frac{-17- \sqrt{409}}{10}$

$\rule{200}{2}$