If the roots of the equation lx 2 + mx + m = 0 are in the ratio p ∶ q, then \(\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} {\rm{\;}} + \sqrt {\frac{m}{l}} \) is equal to
0
Solving Quadratic Roots Ratio Problem
The given equation is a quadratic equation: \(lx^2 + mx + m = 0\).
Let the roots of this equation be \(\alpha\) and \(\beta\). For a standard quadratic equation \(ax^2 + bx + c = 0\), the sum and product of roots are given by:
- Sum of roots: \(\alpha + \beta = -\frac{b}{a}\)
- Product of roots: \(\alpha \beta = \frac{c}{a}\)
In our given equation, \(a=l\), \(b=m\), and \(c=m\). Therefore, the sum and product of roots are:
- Sum of roots: \(\alpha + \beta = -\frac{m}{l}\)
- Product of roots: \(\alpha \beta = \frac{m}{l}\)
We are given that the roots are in the ratio \(p : q\). This means:
\[ \frac{\alpha}{\beta} = \frac{p}{q} \]We need to find the value of the expression \(\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} {\rm{\;}} + \sqrt {\frac{m}{l}}\).
Substitute the ratio of roots into the first two terms of the expression:
\[ \sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} = \sqrt {\frac{\alpha}{\beta}} + \sqrt {\frac{\beta}{\alpha}} \]Combine these two terms by finding a common denominator:
\[ \sqrt {\frac{\alpha}{\beta}} + \sqrt {\frac{\beta}{\alpha}} = \frac{\sqrt{\alpha}}{\sqrt{\beta}} + \frac{\sqrt{\beta}}{\sqrt{\alpha}} = \frac{(\sqrt{\alpha})^2 + (\sqrt{\beta})^2}{\sqrt{\alpha} \sqrt{\beta}} = \frac{\alpha + \beta}{\sqrt{\alpha \beta}} \]Now substitute the expressions for the sum of roots (\(\alpha + \beta\)) and the product of roots (\(\alpha \beta\)) that we found from the quadratic equation:
\[ \frac{\alpha + \beta}{\sqrt{\alpha \beta}} = \frac{-\frac{m}{l}}{\sqrt{\frac{m}{l}}} \]Simplify this expression:
\[ \frac{-\frac{m}{l}}{\sqrt{\frac{m}{l}}} = \frac{-\frac{m}{l}}{\left(\frac{m}{l}\right)^{1/2}} \]Using the property of exponents \(\frac{a^x}{a^y} = a^{x-y}\):
\[ \frac{-\left(\frac{m}{l}\right)^1}{\left(\frac{m}{l}\right)^{1/2}} = - \left(\frac{m}{l}\right)^{1 - \frac{1}{2}} = - \left(\frac{m}{l}\right)^{1/2} = -\sqrt{\frac{m}{l}} \]So, we found that \(\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} = -\sqrt{\frac{m}{l}}\).
Now substitute this result back into the original expression we needed to evaluate:
\[ \left(\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}}\right) + \sqrt {\frac{m}{l}} = \left(-\sqrt{\frac{m}{l}}\right) + \sqrt{\frac{m}{l}} \] \[ = 0 \]Thus, the value of the expression \(\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} {\rm{\;}} + \sqrt {\frac{m}{l}}\) is 0.
Revision Table: Quadratic Roots Properties
| Property | Formula for \(ax^2 + bx + c = 0\) | Formula for \(lx^2 + mx + m = 0\) |
|---|---|---|
| Sum of Roots (\(\alpha + \beta\)) | \(-\frac{b}{a}\) | \(-\frac{m}{l}\) |
| Product of Roots (\(\alpha \beta\)) | \(\frac{c}{a}\) | \(\frac{m}{l}\) |
| Ratio of Roots (\(\alpha : \beta\)) | \(\frac{\alpha}{\beta}\) | \(\frac{p}{q}\) (given) |
Additional Information: Roots in Ratio
When the roots of a quadratic equation \(ax^2 + bx + c = 0\) are in the ratio \(p:q\), we can write the roots as \(k p\) and \(k q\) for some constant \(k\). Using the sum and product of roots:
- Sum: \(kp + kq = k(p+q) = -\frac{b}{a}\)
- Product: \((kp)(kq) = k^2 pq = \frac{c}{a}\)
Dividing the square of the sum by the product gives:
\[ \frac{(k(p+q))^2}{k^2 pq} = \frac{(-b/a)^2}{c/a} \] \[ \frac{k^2 (p+q)^2}{k^2 pq} = \frac{b^2/a^2}{c/a} \] \[ \frac{(p+q)^2}{pq} = \frac{b^2}{a^2} \cdot \frac{a}{c} = \frac{b^2}{ac} \] \[ \frac{p^2 + 2pq + q^2}{pq} = \frac{b^2}{ac} \] \[ \frac{p}{q} + 2 + \frac{q}{p} = \frac{b^2}{ac} \]This identity is useful for problems involving roots in a specific ratio. In this particular problem, we used a direct substitution approach relating the ratio to the roots and then to the sum and product of roots, which also effectively solves the problem.
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 + \(\frac{1}{x^3}\) .
