Question

1. 12x - 40x + 32 = 0
II. 8y- 40y + 48 = 0
quadratic equation​

Answer 1

$\huge{\star}{\underline{\boxed{\red{\sf{Answer :}}}}}{\star}$

Given :-

12x - 40x + 32 = 0 ........(1)

8y - 40y + 48 = 0 .........(2)

================

To Find :-

Value of x and y

================

Solution :-

From equation 1

→ 12x - 40x + 32 = 0

→ -28x + 32 = 0

Taking 32 to RHS

→ - 28x = - 32

→ x = -32/-28

→ x = 8/7

$\huge{\star}{\underline{\boxed{\blue{\sf{x = \frac{8}{7}}}}}}{\star}$

$\rule{200}{2}$

From equation 2.

→ 8y - 40y + 48 = 0

→ -32y + 48 = 0

Take 48 to RHS

→ - 32y = -48

→ y = -48/-32

→ y = 3/2

$\huge{\star}{\underline{\boxed{\green{\sf{y = \frac{3}{2}}}}}}{\star}$

Answer 2

Correct Question

Solve the following quadratic equations :

I) 12x² - 40x + 32 = 0

ll) 8y² - 40y + 48 = 0

Solution

1)

Given Equation,

12x² - 40x + 32 = 0

→3x² - 10x + 8 = 0

By splitting the middle term,

→ 3x² - 6x - 4x + 8 = 0

→ 3x(x - 2) - 4(x - 2) = 0

→ (3x - 4)(x - 2) = 0

→ x = 2 or, 4/3

===========================

2)

Given Equation,

8y² - 40y + 48 = 0

→ y² - 5y + 6 = 0

By splitting the middle term,

→ y² - 2y - 3y + 6 = 0

→ y(y - 2) -3(y - 2) = 0

→ (y - 2)(y - 3) = 0

→ y = 2 or 3

==========================