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dolphi86 [110]
3 years ago
5

What is the number of real solutions? –11x^2 = x + 11

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
8 0
-11 x² = x + 11
11 x² + x + 11 = 0
We will find the value of the discriminant:
D = b² - 4 a c = 1² - 4 * 11 * 11 * 1 - 484 = - 483
D < 0. There are no real solutions.
Aleks04 [339]3 years ago
8 0

Answer:

Given the equation: -11x^2 = x+11

or we can write above equation as:

-11x^2-x-11=0 or

11x^2+x+11 =0

The general equation of quadratic formula for ax^2+bx+c=0; where a, b and c are constant;

Use Discriminant formula: D =b^2-4ac

If D> 0, then there are 2 roots.

If D = 0, then there is only 1 root.

If D <0, then there are no real roots.

Now, from the given equation 11x^2+x+11 =0 we have

a =11, b= 1 and c =11

Then, using discriminant formula we have;

D =b^2-4ac =(1)^2 -4(11)(11) = 1-4\cdot 121 = 1-484= -483

⇒D<0 [ No real roots]

Therefore, for the given equation -11x^2 = x+11 , there are no real solutions.


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