Show that the equation $$x^{4} - 12x^{2} + 12x - 3 = 0$$ has a root between $$-3$$ and $$-4$$ and another between $$2$$ and $$3$$.
Given equation, $$x^4-12x^2+12x-3=0$$
Consider $$f(x) =x^4-12x^2+12x-3$$.
Then $$f(-3) = -66$$ and $$f(-4) = 13$$
Since the signs of $$f(-3)$$ and $$f(-4)$$ are opposite, $$f(x)$$ must cross x-axis atleast once in the interval $$(-4,-3)$$.
$$\therefore\,\, f(x) = 0$$ must have one root between $$-3$$ and $$-4$$.
Similarly, $$f(2) = -11$$ and $$f(3) = 6$$.
Since the signs of $$f(2)$$ and $$f(3)$$ are different, $$f(x) = 0$$ must have one root between $$2$$ and $$3$$.
From the top of a $$64$$ metres high tower, a stone is thrown upwards vertically with a velocity of $$48 \text{ m/s}$$. The greatest height (in metres) attained by the stone, assuming the value of gravitational acceleration $$g = 32 \text{ m/s}^2$$, is (Note: This question was asked in Maths subject in JEE Mains $$2015$$ online exam held on $$11$$ April $$2015$$)
The sum of all the real values of $$x$$ satisfying the equation $$2^{(x-1)(x^2+5x-50)}=1$$ is.
Let $$\alpha$$ and $$\beta$$ be the roots of equation $$x^2-6x-2=0$$. If $$a_n=\alpha^n-\beta^n$$, for $$n\geqslant 1$$, then the value of $$\dfrac {a_{10}-2a_8}{2a_9}$$ is equal to
If $$2x^3+ax^2+bx+4=0$$ (a and b are positive real numbers) has $$3$$ real roots, then prove that $$ a+b \geq 6(2^{1/3}+4^{1/3})$$.
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Find the solutions of the following equations which have common roots: $$4x^{4} + 12x^{3} - x^{2} - 15x = 0, 6x^{4} + 13x^{3} - 4x^{2} - 15x = 0$$.
Find the condition that $$x^{n} - px^{2} + r = 0$$ may have equal roots.
If the equation $$x^{5} - 10a^{3}x^{2} + b^{4}x + c^{5} = 0$$ has three equal roots, show that $$ab^{4} - 9a^{5} + c^{5} = 0$$.
If the roots of the equation $$x^{3} - ax^{2} + x - b = 0$$ are in harmonical progression, show that the mean root is $$3b$$.
$$2$$ and $$-2$$ are two zeros of the polynomial $$m^4+m^3-34m^{2} - 4m + 120$$. What are the other two zeros of the polynomial?