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
If \(a^b = b^a\), then what is
\(\frac{a \times (\frac{a}{b}) ^{ \frac{a}{b} } }{ (a)^{ \frac{a}{b} } }\)
equal to?
If \(x^4 + y^4 = 14x^2 y^2\), then consider the following :
I. \(\log_{10}(x^2 + y^2)\) = \(\log_{10} x + \log_{10} y + 2\log_{10} 2\)
II. \(\log_{10}(x^2-y^2)=\log_{10} x + \log_{10} y\) +\(\log_{10}2+0.5\log_{10} 3\)
Which of the above is/are correct?
- If \(\log_{10}2 = 0.301\) and \(\log_{10} 3 = 0.477\), then what is the number of digits in the expansion of \(60^{60}\) ?
- If \(\log_{10}(80)=p\), \(\log_{10}(45)=q\) and \(\log_{10}(216)=r\), then what is \(\log_{10}(384)\) equal to ?
If log10(100001 - 4x)/(5 - x) = 1, then what is x equal to?
What is √17 - 4√15 + √8 - 2√15 equal to?
A sum of money at the rate of \(5\%\) per annum compounded annually becomes \(n\) times in \(100\) years. What is the value of \(n\)? (Given \(\log_{10}2=0.301\), \(\log_{10}3=0.477\) and \(\log_{10}7=0.845\))
What is the number of zeros immediately after the decimal point in \((0.5)^{1000}\)? (Given that \(\log_{10}2 = 0.30103\))
Consider the following statements in respect of common logarithms :
I. The logarithm of a number greater than 100 but less than 1000 lies between 2 and 3.
II. The logarithm of a positive number less than unity is negative.
Which of the statements given above is/are correct?
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:
- If \(p + q = 15\), then what is \(q-p\) equal to?
- If \(p + q = 66\), then which one of the following is correct?
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 ?
