Question

hey can you please help me posted picture of question :)

Answer 1

-x^2+5=(4x-1)^2
Using in the right side of the equation: (a-b)^2=a^2-2ab+b^2, with a=4x and b=1:
-x^2+5=(4x)^2-2(4x)(1)+(1)^2
-x^2+5=16x^2-8x+1
-x^2+5+x^2-5=16x^2-8x+1+x^2-5
0=17x^2-8x-4
17x^2-8x-4=0

Answer: Option D. 17x^2-8x-4=0

Answer 2

Answer:
17x² - 8x - 4 = 0

Explanation:
First, we would need to expand the bracket on the right:
(4x-1)² = (4x)² - 2(4x)(1) + (1)²
(4x-1)² = 16x² - 8x + 1

The given equation now becomes:
-x² + 5 = 16x² - 8x + 1

Now, the standard form of the quadratic equation is:
ax² + bx + c = 0

This means that we want all the terms on one side and a zero on the otehr.
Therefore:
-x² + 5 = 16x² - 8x + 1
-x² + 5 - 16x² + 8x - 1 = 0
-17x² + 8x + 4 = 0
Multiply all terms by -1 to get a positive value for the coefficient of the x².
This will give:
17x² - 8x - 4 = 0

Hope this helps :)