Solutions for the provided equation x² + 8x + 16 = 7 are
Either x = -4 + √7, or x = -4 - √7
∴ Options A and F are applicable to the given equation.
To solve a quadratic equation we often use a formula which is known as Sridharacharya Formula. The formula for the standard equation
ax² + bx + c = 0 is given by,
x = (-b±√(b² - 4ac))/2a
We have been given,
x² + 8x + 16 = 7
or, x² + 8x + 16 - 7 = 0
or, x² + 8x + 9 = 0.
This equation is now in the standard form, so we can apply Sridharacharya Formula taking a=1, b=8, c=9.
∴ x = (-8±√(8² - 4*1*9))/2*1 = (-8±√(64 - 36))/2 = (-8±√28)/2 = (-8 ± 2√7)/2
⇒ x = -4 ± √7,
∴ Either x = -4 + √7, or x = -4 - √7
Among the given options, Option A. and Option F. satisfies our equation.
Learn more about Sridharacharya Formula at
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