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21). If one roots of the equation is reciprocal of the other, then which one of the following is correct?
Given quadratic equation is | ||||

22). Number of solutions of the equation \( \Large \sqrt{x^{2}-x+1}+\frac{1}{\sqrt{x^{2}-x+1}}=2-x^{2} \) is
We know that, \( \Large AM \ge GM \) | ||||

23). If \( \Large x = \sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}} \), then \( \Large x^{2}-x-1 \) is equal to
Here, x = \( \Large \sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}} \) On rationalising the terms given root, we get x = \( \Large \sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}-1} } = \frac{\sqrt{5}+1}{2} \) Now, substituting the value of x in \( \Large x^{2} - x - 1. \) Therefore, \( \Large x^{2} -x - 1 = \left(\frac{\sqrt{5}+1}{2}\right)^{2} - \left(\frac{\sqrt{5}+1}{2}\right) - 1 \) = \( \Large \frac{5+1+2\sqrt{5}}{4} - \frac{\sqrt{5}+1}{2} - 1 \) = \( \Large \frac{6 + 2 \sqrt{5} - 2 \sqrt{5} - 2 - 4}{4} = 0 \) | ||||

24). If one of the roots of the equation \( \Large x^{2}-bx+c=0 \) is the square of the other, then which of the following option is correct?
Given that, one root of the equation \( \Large x^{2} - bx + c = 0\) is square of other root of this equation i.e., roots (\( \Large \left( \alpha , \alpha ^{2}\right) \)). | ||||

25). Two students A and B solve an equation of the form \( \Large x^{2}+px+q=0 \). A starts with a wrong value of p and obtains the roots as 2 and 6. B starts with a wrong value of q and gets the roots as 2 and -9. What are the correct roots of the equation?
Let \( \Large \alpha \) and \( \Large \beta \) be the roots of the | ||||

26). If the roots of equations \( \Large ax^{2}+bx+c=0 \) be \( \Large \ \alpha \ and\ \beta \) then the roots of equations \( \Large cx^{2}+bx+a=0 \) are:
Since \( \Large \alpha \) and \( \Large \beta \) are the roots by \( \Large ax^{2}+bx+c=0 \) => \( \Large \alpha + \beta \) = \( \Large -\frac{b}{c} \) and \( \Large \alpha \beta = \frac{c}{a} \) | ||||

27). Both the roots of the given equation. \( \Large \left(x-a\right) \left(x-b\right)+ \left(x-b\right) \left(x-c\right)+ \left(x-c\right) \left(x-a\right)=0 \) are always
Given equation: \( \Large \left(x-a\right) \left(x-b\right)+ \left(x-b\right) \left(x-c\right)+ \left(x-c\right) \left(x-a\right)=0 \) \( \Large 3x^{2}-2 \left(a+b+c\right)x+ \left(ab+bc+ca\right)=0 \) Now, \( \Large BZ-4AC=4\left[ \left(a+b+c+\right)^{2}-3 \left(ab+bc+ca\right) \right] \) = \( \Large 4 \left(a^{2}+b^{2}+c^{2}-ab-bc-ac\right) \) = \( \Large 2\{ \left(a-b\right)^{2}+ \left(b-c\right)^{2}+ \left(c-a\right)^{2} \}\ge 0 \) Hence, both roots are always real. | ||||

28). If the difference of roots of the equation \( \Large x^{2}-bx+c=0 \) be 1, then;
Let \( \Large \alpha \) and \( \Large \beta \) are the roots of equation \( \Large x^{2}-bx+c=0 \) => \( \Large \alpha + \beta = b\ and\ \alpha \beta = c\) => \( \Large \alpha + \beta = \sqrt{ \left( \alpha + \beta \right)^{2}-4 \alpha \beta } \) => \( \Large 1 = \sqrt{b^{2}- 4c} \) => \( \Large b^{2}-4c-1=0 \) | ||||

29). If \( \Large 2+i\sqrt{3} \) is a root of the equation \( \Large x^{2}+px+q=0 \) where p and q are real, then \( \Large p, q \) is equal to:
Since, \( \Large 2+i\sqrt{3} \) is a root of equation \( \Large x^{2}+px+q=0 \) therefore, \( \Large 2-i\sqrt{3} \) will be other root. Now sum of the roots \( \Large 2+i\sqrt{3} \) + \( \Large 2-i\sqrt{3} \) = -P | ||||

30). If \( \Large x=\sqrt{1+\sqrt{1+\sqrt{1}+.....}} \), the x is equal to
We have, \( \Large x=\sqrt{1+\sqrt{1+\sqrt{1+.....\infty}}} \) |