Q.
it is the question on linear equation in one variable
It is algebra
the answer for this question is
I have given the answer
but now please solve me
please solve this question for me don't give me any rubbish answer if you don't know
give me correct step by step method and correct answer
Answers
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Step-by-step explanation:
(x - 5 ) ² + ( x + 2 ) ² = - 2
or, x² - 10 x + 25 + x² + 4x + 4 = - 2
{ Using, (a -b)² = a² - 2ab + b²
and (a + b)² = a² + 2ab + b² }
or, 2x² - 6x + 29 + 2 = 0
( Solving, and transposing - 2 to LHS )
or, 2x² - 6x + 31 = 0
We solve it using Quadratic formula,
Here, a = 2 , b = - 6 , c = 31
Discriminant, D = b² - 4ac
= (- 6)² - 4 × 2 × 31
= 36 - 248
= - 212, which is less than zero
Thus, no real roots exist for the given Quadratic Equation.
Plz mark as brainliest.
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