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4 years ago

Could someone please help me with this? Solution set of lx^2+mx+n=0 is...?

Mathematics

1 answer:

IgorLugansk [536]4 years ago

6 0

Answer:

Simplifying

lx2 + mx + n = 0

Solving

lx2 + mx + n = 0

Solving for variable 'l'.

Move all terms containing l to the left, all other terms to the right.

Add '-1mx' to each side of the equation.

lx2 + mx + -1mx + n = 0 + -1mx

Combine like terms: mx + -1mx = 0

lx2 + 0 + n = 0 + -1mx

lx2 + n = 0 + -1mx

Remove the zero:

lx2 + n = -1mx

Add '-1n' to each side of the equation.

lx2 + n + -1n = -1mx + -1n

Combine like terms: n + -1n = 0

lx2 + 0 = -1mx + -1n

lx2 = -1mx + -1n

Divide each side by 'x2'.

l = -1mx-1 + -1nx-2

Simplifying

l = -1mx-1 + -1nx-2

Step-by-step explanation:

Hope this helped you!

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Complete Question

The complete question is shown on the first uploaded image

Answer:

$g(h(x)) = \sqrt[4]{x^2 + x +5} + 3$

$h(g(x)) = \sqrt{x} + 7\sqrt[4]{x} + 17$

$h(h(x)) = [x^2 + x + 5 ]^2 + x^2 + x + 10$

Step-by-step explanation:

From the question we are told that

$h(x) = x^2 + x + 5$

and

$g(x) = \sqrt[4]{x} + 3$

Considering first question

Now we are told g(h(x))

i.e

$g(h(x)) = [x^2 + x + 5 ]^{\frac{1}{4} } + 3$

=> $g(h(x)) = \sqrt[4]{x^2 + x +5} + 3$

Considering second question

Now we are told h(g(x))

i.e

$h(g(x)) = [x^{\frac{1}{4} } + 3]^2 + x^{\frac{1}{4} } + 3 + 5$

=> $h(g(x)) = x^{\frac{1}{2} } + 6x^{\frac{1}{4} } + 9+ x^{\frac{1}{4}} + 8$

=> $h(g(x)) = x^{\frac{1}{2}} + 7x^{\frac{1}{4}} + 17$

=> $h(g(x)) = \sqrt{x} + 7\sqrt[4]{x} + 17$

Considering third question

$h(h(x))= [x^2 + x + 5]^2 + [x^2 + x + 5 ] + 5$

=> $h(h(x)) = [x^2 + x + 5 ]^2 + x^2 + x + 10$

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