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jeka94
3 years ago
14

Write 9− 16x⁴ as the product of two factors.

Mathematics
1 answer:
Marysya12 [62]3 years ago
6 0

Answer:

The two factors will be =(3+4x^2)\ and\ (3-4x^2)        

Step-by-step explanation:

We have given the equation 9-16x^4

We know the algebraic equation a^2-b^2=(a+b)(a-b)

As  9-16x^4 can be written as 3^2-(4x^2)^2, as square of 3 is 9 and square of 4x^2 is 16x^4

So 9-16x^4=3^2-(4x^2)^=(3+4x^2)(3-4x^2)

So the two factors will be =(3+4x^2)\ and\ (3-4x^2)

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1) Determine the discriminant of the 2nd degree equation below:
Aleksandr-060686 [28]

\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}

We have, Discriminant formula for finding roots:

\large{ \boxed{ \rm{x =  \frac{  - b \pm \:  \sqrt{ {b}^{2}  - 4ac} }{2a} }}}

Here,

  • x is the root of the equation.
  • a is the coefficient of x^2
  • b is the coefficient of x
  • c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5\pm  \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5  \pm  \sqrt{25 - 24} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5 \pm 1}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x =  - 2 \: or  - 3}}}

❒ p(x) = x^2 + 2x + 1 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{  - 2 \pm  \sqrt{ {2}^{2}  - 4 \times 1 \times 1} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 2 \pm 0}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x =  - 1 \: or \:  - 1}}}

❒ p(x) = x^2 - x - 20 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - ( - 1) \pm  \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{1 \pm 9}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \:  - 4}}}

❒ p(x) = x^2 - 3x - 4 = 0

\large{ \rm{ \longrightarrow \: x =   \dfrac{  - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{3 \pm \sqrt{9  + 16} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{3  \pm 5}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \:  - 1}}}

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