Question

40x=10x^2+41 Identify the number of solutions and their types using the discriminant

Answer 1

Number of solutions: no roots

Type of solution: Not Real

Step-by-step explanation:

We need to identify the number of solutions and their types using the discriminant.

We are given: $40x=10x^2+41$

Rearranging:

$40x-10x^2-41=0\\10x^2-40x+41=0$

Discriminant can be found by: $b^2-4ac$

where b=-40, a=10 and c=41

Putting values:

$b^2-4ac\\=(-40)^2-4(10)(41)\\=1600-1640\\=-40$

So, Discriminant is -40

If the discriminant is less than zero i.e -40 then there are no real roots.

So, Number of solutions: no roots

Type of solution: Not Real

Keywords: discriminant

Learn more about discriminant at:

• brainly.com/question/8196933
• brainly.com/question/9328925
• brainly.com/question/9184197

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