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Answer : Option C
x = ±√5, x = ±10iExplanation :
Given equation,
x^4 + 95x^2 – 500 = 0By factoring the above equation we get,
x^4 + 100x^2 – 5x^2 – 500 = 0
x^2(x^2 + 100) – 5(x^2 + 100) = 0
(x^2 – 5)(x^2 + 100) = 0solving the equation,
x^2 = 5, x^2 = 100Therefore, x = ±√5, x = ±10i
Solution: To find the inverse equation, interchange x and y variables,
x = 100 – y²
Then find roots of y,
y² = 100 – x
y = ±√(100 – x)
Therefore, the inverse equation for y = 100 – x^2
is, y = ±√(100 – x)Answer : Option D
Explanation :
=> y = 3(x – 2)^2 – (x – 5)^2
=> y = 3(x – 2)(x – 2) – (x – 5)(x – 5)
=> y = 3(x(x – 2) – 2(x – 2)) – (x(x – 5) – 5(x – 5))
=> y = 3(x(x) – x(2) – 2(x) + 2(2)) – (x(x) – x(5) – 5(x) + 5(5))=> y = 3(x^2 – 2x – 2x + 4) – (x^2 – 5x – 5x + 25)
=> y = 3(x^2 – 4x + 4) – (x^2 – 10x + 25)
=> y = 3(x^2) – 3(4x) + 3(4) – (x^2) + (10x) – (25)
=> y = (3x^2 – 12x + 12) + (-x^2 + 10x – 25)
=> y = (3x^2 – x^2) + (-12x + 10x) + (12 – 25)
=> y = 2x^2 – 2x – 13add 13 on both sides
=> y + 13 = 2x^2 – 2x + 1/2y + 13 + 1/2 = 2(x^2 – x + 1/4)
=> y + 27/2 = 2(x^2 – 1/2x – 1/2x + 1/4)
=> y + 27/2 = 2(x(x) – x(0.5) – 0.5(x) + 0.5(0.5))
=> y + 27/2 = 2(x(x – 1/2) – 1/2(x – 1/2))
=> y + 27/2 = 2(x – 1/2)(x – 1/2)
=> y + 27/2 = 2(x – 1/2)^2subtract 13.5 on both sides
Hence, y = 2(x – 1/2)^2 – 27/2
The Slope of line, y = x – 3 is 1.
Explanation :
the standard equation of the line in the slope-y intercept form is y = m.x + c
where,
m is slope of the line
c is the y-intercept of lineBy comparing the above equation with standard form
y = (1)x – 3Therefore, the Slope of line, y = x – 3 is 1.
Answer: x = (2 ± √10 i)/2
Explanation :
Standard Quadratic Expression : ax^2 + bx + c = 0
Using quadratic formula, expression for roots,
x = [ -b ± √(b2 – 4ac) ] / 2aBy comparing,
=> x = [ -(-4) ± √{(-4)2 – 4(2)(7)} ] / 2(2)
=> x = [ 4 ± √{16 – 56} ] / 4
=> x = [ 4 ± √(-40) ] / 4
=> x = [ 4 ± 2√10 i ] / 4
=> x = (2 ± √10i)/2Therefore, the roots of x are : (2 + √10 i)/2 and (2 – √10 i)/2