Solving the Equation $9(x+9)^2=441$
To find the roots of the given equation, we first isolate the squared term:
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Divide both sides by 9:
$ (x+9)^2 = \frac{441}{9} $
$ (x+9)^2 = 49 $
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Take the square root of both sides. Remember to include both positive and negative roots:
$ x+9 = \pm\sqrt{49} $
$ x+9 = \pm 7 $
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Solve for x in both cases:
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Case 1 (Positive root):
$ x + 9 = 7 $
$ x = 7 - 9 $
$ x = -2 $
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Case 2 (Negative root):
$ x + 9 = -7 $
$ x = -7 - 9 $
$ x = -16 $
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Therefore, the roots of the equation $9(x+9)^2=441$ are -2 and -16.
Verification
The roots found are -2 and -16, which corresponds to Option D.
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Important Questions from Quadratic Equation
If 2x 2+ 5x + 1 = 0, then one of the values of \(x - \frac{1}{{2x}}\) is:
If \(a-\frac{12}{a}=1\) , where a > 0, then the value of \(a^2+\frac{16}{a^2}\) is:
If x 2 – 3x + 1 = 0, then the value of \(\frac{(x^4+\frac{1}{x^2})}{(x^2+5x+1)}\) is:
If \(\sqrt{x}{}-{1\over\sqrt{x}}=\sqrt5\) , \(x \ne 0\) , then what is the value of \((x^4+{1\over{x^2}})\over(x^2+1) \) ?
If x 2 + \(\frac{1}{x^2}\) = 18, x > 0, then find the value of x 3 + \(\frac{1}{x^3}\) .
