A2b2x2 - 4b2 - 3a4 x - 12a2b2 - 0- a- 0 and b- 0-

a ^ 2 b ^ 2 x ^ 2 ( 4 b ^ 2 3 a ^ 4 ) x 12 a ^ 2 b ^ 2 =0 Therefore, the roots are: x=( 4 b ^ 2 3 a ^ 4 ) ±√(( 4 b ^ 2 3 a ^ 4 ) #160; ^ 2 +4 ...

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Standard X

Mathematics

Quadratic Formula

a2b2x2 - 4b2 ...

Question

$a^{2} b^{2} x^{2} - (4 b^{2} - 3 a^{4}) x - 12 a^{2} b^{2} = 0, a \neq 0$ and $b \neq 0$ .

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Solution

$a^{2} b^{2} x^{2} - (4 b^{2} - 3 a^{4}) x - 12 a^{2} b^{2} = 0$
Therefore, the roots are:
$x = \frac{(4 b^{2} - 3 a^{4}) \pm \sqrt{{(4 b^{2} - 3 a^{4})}^{2} + 4 a^{2} b^{2} (12 a^{2} b^{2})}}{2 a^{2} b^{2}} = \frac{(4 b^{2} - 3 a^{4}) \pm \sqrt{9 a^{8} + 16 b^{4} - 24 a^{4} b^{2} + 48 a^{4} b^{4}}}{2 a^{2} b^{2}} = \frac{(4 b^{2} - 3 a^{4}) \pm \sqrt{{(4 b^{2} + 3 a^{4})}^{2}}}{2 a^{2} b^{2}} = \frac{(4 b^{2} - 3 a^{4}) \pm (4 b^{2} + 3 a^{4})}{2 a^{2} b^{2}} ∴ x = \frac{- 6 a^{4}}{2 a^{2} b^{2}} = \frac{- 3 a^{2}}{b^{2}} o r, x = \frac{8 b^{2}}{2 a^{2} b^{2}} = \frac{4}{a^{2}}$

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Similar questions

using quadratic formula solve for x : a²b²x² - (4b⁴ - 3a⁴) x - 12a²b² = 0

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
$a^{2} b^{2} x^{2} - (4 b^{4} - 3 a^{4}) x - 12 a^{2} b^{2} = 0, a \neq 0 a n d b \neq 0$

Q. Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

a^{2} b^{2} x^{2} - (4 b^{4} - 3 a^{4}) x - 12 a^{2} b^{2} = 0

a \neq 0

and

b \neq 0

[CBSE 2006]

a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0

Q. Solve:

a^{2} b^{2} x^{2} - (4 b^{4} - 3 a^{4}) x - 12 a^{2} b^{2} = 0 .

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a2b2x2−(4b2−3a4)x−12a2b2=0,a≠0 and b≠0.

a2b2x2−(4b2−3a4)x−12a2b2=0Therefore, the roots are:x=(4b2−3a4)±√(4b2−3a4)2+4a2b2(12a2b2)2a2b2=(4b2−3a4)±√9a8+16b4−24a4b2+48a4b42a2b2=(4b2−3a4)±√(4b2+3a4)22a2b2=(4b2−3a4)±(4b2+3a4)2a2b2∴x=−6a42a2b2=−3a2b2or,x=8b22a2b2=4a2

$a^{2} b^{2} x^{2} - (4 b^{2} - 3 a^{4}) x - 12 a^{2} b^{2} = 0, a \neq 0$ and $b \neq 0$ .