Question

pls solve the sum of qudratic equation step by step​

Answer 1

Hello friend..

Answer:

Step-by-step explanation:

Let the larger part be x . Then the smaller part is = 16-x

So according to the question..

2x^2 = (16-x)^2 +164

2x^2-(16-x)^2-164=0

By solving this equation we get

x^2+32x-420 =0

By factorization we get..

(x+42) (x-10) =0

x= -42 or x =10

x=10. { x = -42 is not possible}

Hope it helps..✌✌

Answer 2

Answer:

Larger part = 10, smaller part = 6

Step-by-step explanation:

Let the first part (larger) be 'X'

The the second part (smaller) is '16-X'

Given, Twice the square of larger exceeds square of smaller by 164

2 X^2 - (16 - X)^2 = 164

2X^2 - (256 - 32X + X^2) = 164

2X^2 - 256 + 32X - X^2 = 164

2X^2 + 32 X - 256 - 164 = 0

X^2 + 32X - 420 = 0

Sum = 32 && Product = -420

Numbers: 42,-10

X^2 -10X + 42X - 420 = 0

X(X-10) + 42(X-10) = 0

[X-10][X+42] = 0

X = 10

Number divided as : 10 , 6