Question

expand 4x² - 1 - (7x²+13x-2)

Answer 1

Step-by-step explanation:

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☆☞ [ Verified answer ]☜☆

answer is 4x²-1-4x

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Answer 2

Solution :-

→ 4x² - 1 - (7x² + 13x - 2)

→ 4x² - 1 - 7x² - 13x + 2

→ 4x² - 7x² - 13x + 2 - 1

→ (-3)x² - 13x + 1

Now,

if we want to solve for x ,

put the Given Equation to 0 .

Sridharacharya formula for Solving Quadratic Equation ax² +bx + c = 0 says that,

→ x = [ -b±√(b²-4ac) / 2a ]

or,

→ x = [ - b± √D /2a ] where D(Discriminant)= b²-4ac.

So,

→ (-3)x² - 13x + 1 = 0

→ (-1)[3x² + 13x - 1] = 0

→ 3x² + 13x - 1 = 0

comparing with ax² + bx + c = 0, we get,

• a = 3
• b = 13
• c = (-1)

Than,

→ D = b² - 4ac

→ D = (13)² - 4*3*(-1)

→ D = 169 + 12

→ D = 181 .

Therefore,

→ x = [ - b± √D /2a ]

→ x = (- 13 ± √181) / 2*3

→ x = (- 13 ± √181) / 6

→ x = (-13 + √181) / 6 or, (-13 - √181) / 6 .