To solve the problem of finding \((\alpha - \beta)^2\) where \(\alpha\) and \(\beta\) are roots of the equation \(\log_{10} [998+\sqrt{x^2-18x+76}] = 3\), let's follow these steps:
- Start with solving the logarithmic equation: \(\log_{10} [998+\sqrt{x^2-18x+76}] = 3\).
- Convert the logarithmic form to exponential form: \(998+\sqrt{x^2-18x+76} = 10^3\).
- Calculate the power of 10: \(10^3 = 1000\).
- Substitute back to find the equation: \(998+\sqrt{x^2-18x+76} = 1000\) leads to \(\sqrt{x^2-18x+76} = 2\).
- Square both sides to eliminate the square root: \(x^2 - 18x + 76 = 4\).
- Reorganize the equation into a standard quadratic form: \(x^2 - 18x + 72 = 0\).
- For a quadratic equation of the form \(ax^2 + bx + c = 0\), the roots satisfy \(\alpha + \beta = -\frac{b}{a}\) and \(\alpha\beta = \frac{c}{a}\). Here, \(a = 1, b = -18, c = 72\).
- Calculate \(\alpha + \beta = 18\) and \(\alpha \beta = 72\).
- Use the identity for the square of the difference of roots: \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\).
- Substitute the known values:
- \((\alpha + \beta)^2 = 18^2 = 324\)
- \(4\alpha\beta = 4 \times 72 = 288\)
- Calculate: \((\alpha - \beta)^2 = 324 - 288 = 36\).
The value of \((\alpha - \beta)^2\) is therefore 36.
Similar Questions
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\(\frac{a \times (\frac{a}{b}) ^{ \frac{a}{b} } }{ (a)^{ \frac{a}{b} } }\)
equal to?
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Which of the above is/are correct?
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Important Questions from Logarithms
Given that $132^{0.14} = x$, $132^{0.26} = y$ and $x^z = y^2$, then the value of z is close to:
For \(x \ge y > 1\), let \(\log_x\left(\frac{x}{y}\right) + \log_y\left(\frac{y}{x}\right) = k\), then the value of \(k\) can never be equal to
If \(\log_ba = p\), \(\log_dc = 2p\) and \(\log_fe = 3p\), then what is \((ace)^{\frac{1}{p}}\) equal to ?
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- Let \(p = \ln(x)\), \(q = \ln(x^3)\) and \(r = \ln(x^5)\), where \(x > 1\). Which of the following statements is/are correct?
I. \(p, q\) and \(r\) are in AP.
II. \(p, q\) and \(r\) can never be in GP.
Select the answer using the code given below.
