13.) x^4-10x^2+24=0 --- (14.) x^4*-10x^2+ 21=0 --- (15.) x^4+54=15x^2?

Question

I need help solve!
use variable substitution and factoring to find all of the roots of each equation. If necessary, leave your answers in radical form.

Answer 1

13.) x^4-10x^2+24=0
Let y = x^2, then we have
y^2 - 10y + 24 = 0
<=> (y - 6)(y - 4) = 0
<=> y = 4 or 6
<=> x^2 = 4 or 6
<=> x = ±2, ±√6.

14.) x^4-10x^2+ 21=0
Exactly the same: let y = x^2, then we have
y^2 - 10y + 21 = 0
<=> (y - 3)(y - 7) = 0
<=> y = 3 or 7
<=> x^2 = 3 or 7
<=> x = ±√3 or ±√7.

15.) x^4+54=15x^2
Same again: let y = x^2, then we have
y^2 + 54 = 15y
<=> y^2 - 15y + 54 = 0
<=> (y - 6)(y - 9) = 0
<=> y = 6 or 9
<=> x^2 = 6 or 9
<=> x = ±√6 or ±3.

Answer 2

x⁴ − x³ + 10x² − 4x + 24 = 0 Since 2i is a root of the above equation, then so is its complex conjugate −2i, otherwise, we would have complex coefficients. So (x − 2i) (x + 2i) are factors of x⁴ − x³ + 10x² − 4x + 24 (x⁴ − x³ + 10x² − 4x + 24) / ((x − 2i) (x + 2i)) = (x⁴ − x³ + 10x² − 4x + 24) / (x² − 4i²) = (x⁴ − x³ + 10x² − 4x + 24) / (x² + 4) = x² − x + 6 x⁴ − x³ + 10x² − 4x + 24 = 0 (x² + 4) (x² − x + 6) = 0 x² + 4 = 0 x = ± 2i x² − x + 6 x = (1 ± √(1−24)) / 2 x = (1 ± √23 i) / 2

Answer 3

13) x^4-10x^2+24=0
Let y = x^2
y^2 - 10y + 24 = 0
By factorisation: (y-4)(y-6)=0
=> y = 4 or y = 6
=> x^2 = 4 or x^2 = 6
=> x=4^(1/2) or x=6^(1/2)
=> x=+/-2 or x= +/-2.449 (corrected to 3 decimal places)

14) x^4*-10x^2+ 21=0
Let y = x^2
y^2 - 10y + 21 = 0
By factorisation: (y-3)(y-7)=0
=> y = 3 or y = 7
=> x^2 = 3 or x^2 = 7
=> x=3^(1/2) or x=7^(1/2)
=> x=+/-1.732 or x= +/-2.6458 (both answers corrected to 3 decimal places)

15) x^4+54=15x^2
x^4 - 15x^2 + 54 = 0
Let y = x^2
y^2 - 15y + 54 = 0
By factorisation: (y-9)(y-6)=0
=> y = 9 or y = 6
=> x^2 = 9 or x^2 = 6
=> x=9^(1/2) or x=6^(1/2)
=> x=+/-3 or x= +/-2.449 (corrected to 3 decimal places)

Answer 4

13. (x^2-4)(x^2-6)=0
x=+-2, +-sqrt6
14. (x^2-3)(x^2-7)=0
x=+-sqrt3, +-sqrt7
15. (x^2-6)(x^2-9)=0
x=+-sqrt6, +-3

Answer 5

#13 (x^2-6)(x^2-4)
x = +/- square root of 6
x = +/- 2

#14 (x^2 -7)(x^2-3) =0
x = +/- square root of 7
x = +/- square root of 3

#15 (x^2-7)(x^2 -8) = 0
x = +/- square root of 7
x = +/- 2*square root of 2

Answer 6

use the quadratic formula!
-b plus or minus sqrt b^2-4ac/2
10+or- sqrt 100-4(24)/2
10+or-sqrt36/2
10+or-3
x=7 or x=13

do the same for the rest and remember, you can't have radicals in the denominator!