Let f(x) = x^2 − 5. Find f(-3)

Mathematics

2 answers:

Art [367]2 years ago

8 0

Answer:

f(-3) = 4

Step-by-step explanation:

f(x) = x² - 5

f(x) is a function with input x, meaning it is an expression which has terms involving x;

x is a variable and can be considered a space holder;

f(-3) is the function with input -3, this simply means replacing the space holder, i.e. x, with -3;

f(-3) = (-3)² - 5

f(-3) = 9 - 5

f(-3) = 4

Art [367]2 years ago

5 0

Answer:

f(-3)=4

Step-by-step explanation:

Substitute x=-3 into the function

f(-3)=(-3)^2-5

f(-3)=9-5

f(-3)=4

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Solve for x:a(a²+b²)x²+b²x-a

m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}$

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}$

$\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}$