4 x square - 3 x + 7
MridulAhi:
what do we need to do?
Factorisation of this polynomial is not possible as this polynomial has imaginary roots
In which class r u?
He is right @snehitha real factorisation can't possible . becoz discriminant= 9-4×7×4 <0
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4x² - 3x + 7
we know , any quadratic equation have real roots only if discriminant ≥ 0
but here, D = b² - 4ac = (-3)² -4×7×4 < 0
so, real roots can't be possible .but imaginary roots possible .
now,
4x² - 3x + 7 = 0
divide both sides by 4
x² - (3/4)x + 7/4 = 0
x² - (3/4)x = (-7/4)
add (3/8)² both sides,
x² - (3/4)x + (3/8)² = (3/8)² - 7/4
(x - 3/8)² = 9/64 - 7/4 = (9 - 112)/64
(x - 3/8)² = -103/64
take square root both sides,
(x - 3/8) = ±√(-103)/8
x - 38 = ±√103 i/8 [ √(-1) = i (from complex no ]
x = 3/8 ± √103i/8
hence, (3 + i√103)/8 and (3 -i√103)/8 are the roots of 4x² -3x + 7
{x - (3+i√103)/8} and { x -(3-i√103)/8 } are the factors of 4x² -3x + 7
we know , any quadratic equation have real roots only if discriminant ≥ 0
but here, D = b² - 4ac = (-3)² -4×7×4 < 0
so, real roots can't be possible .but imaginary roots possible .
now,
4x² - 3x + 7 = 0
divide both sides by 4
x² - (3/4)x + 7/4 = 0
x² - (3/4)x = (-7/4)
add (3/8)² both sides,
x² - (3/4)x + (3/8)² = (3/8)² - 7/4
(x - 3/8)² = 9/64 - 7/4 = (9 - 112)/64
(x - 3/8)² = -103/64
take square root both sides,
(x - 3/8) = ±√(-103)/8
x - 38 = ±√103 i/8 [ √(-1) = i (from complex no ]
x = 3/8 ± √103i/8
hence, (3 + i√103)/8 and (3 -i√103)/8 are the roots of 4x² -3x + 7
{x - (3+i√103)/8} and { x -(3-i√103)/8 } are the factors of 4x² -3x + 7
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