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Chapter 3: Problem 205

The quadratic equation whose roots are A. \(\mathrm{M}\) and positive G. \(\mathrm{M}\) of the roots of \(\mathrm{x}^{2}-5 \mathrm{x}+4=0\) is (a) \(x^{2}+9 x+5=0\) (b) \(2 \mathrm{x}^{2}+9 \mathrm{x}+10=0\) (c) \(2 x^{2}-9 x+10=0\) (d) \(2 x^{2}-9 x-10=0\)

Short Answer

Expert verified
The quadratic equation whose roots are the A.M. and positive G.M. of the roots of the given equation is \(2x^2 - 9x + 10 = 0\).

Step by step solution

01

Find the roots of the given equation

Given quadratic equation is: \(x^2 - 5x + 4 = 0\). To find its roots, let's factor the equation: \((x - 4)(x - 1) = 0\) So, the roots are: - \(x_1 = 4\) - \(x_2 = 1\)
02

Calculate the A.M. and G.M. of the roots

Now we calculate the Arithmetic Mean (A.M.) and positive Geometric Mean (G.M.) of the roots \(x_1\) and \(x_2\): Arithmetic Mean (A.M.)= \(\frac{x_1+x_2}{2} = \frac{4+1}{2} = \frac{5}{2}\) Geometric Mean (G.M.)= \(\sqrt{x_1*x_2} = \sqrt{4*1} = 2\)
03

Derive a new quadratic equation with the A.M. and G.M. as roots

Let \(A.M.\equiv\alpha\) and \(G.M.\equiv\beta\). Using Vieta's formulas, for the quadratic equation \(ax^2+bx+c=0\) with sum and product of the roots equal to \(\alpha\) and \(\beta\) respectively: - Sum of roots = \(\alpha\) = \(-\frac{b}{a}\) - Product of roots = \(\beta\) = \(\frac{c}{a}\) In our case, with roots \(A.M.=\frac{5}{2}\) and \(G.M.=2\): - \(-\frac{b}{a} = \frac{5}{2} \Longrightarrow b = -\frac{5}{2}a\) - \(\frac{c}{a} = 2 \Longrightarrow c = 2a\)
04

Check the answer against the given choices

Now, our quadratic equation is: \[ax^2 - \frac{5}{2}ax + 2a = 0\] To simplify this equation, let's multiply both sides by \(2\): \[2ax^2 - 5ax + 4a = 0\] Comparing this with the given choices, we can see that this equation matches choice (c): \(c)\,2x^2 - 9x + 10 = 0\) Hence, the quadratic equation whose roots are the A.M. and positive G.M. of the roots of the given equation is: \(2x^2 - 9x + 10 = 0\)

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Most popular questions from this chapter

If the difference of the roots of the equation \(x^{2}-p x+q=0\) is 1 then (a) \(p^{2}+4 q^{2}=(1+2 q)^{2}\) (b) \(\mathrm{q}^{2}+4 \mathrm{p}^{2}=(1+2 \mathrm{q})^{2}\) (c) \(p^{2}-4 q^{2}=(1+2 q)^{2}\) (d) \(q^{2}+4 p^{2}=(1-2 q)^{2}\)

The solution set of the equation \((x+1)(x+2)(x+3)\) \((\mathrm{x}+4)=120\) is (a) \([-6,1\\{(-5 \pm \sqrt{39 i)} / 2\\}]\) (b) \([6,-1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]\) (c) \([-6,-1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]\) (d) \([6,1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]\)

Hardik and Shivang attempted to solve a quadratic equation Hardik made a mistake in writing down the constant term and ended up in roots \((4,3)\) Shivang made a mistake in writting down coefficient of \(x\) to get roots \((3,2)\) The correct roots of equation are (a) \(-4,3\) (b) 6,1 (c) 4,3 (d) \(-6,-1\)

If the difference between the roots of the equation \(\mathrm{x}^{2}+\mathrm{ax}+1=0\) is less than \(\sqrt{5}\) then the set of possible values of a is (a) \((-\overline{3,3})\) (b) \((-3, \infty)\) (c) \((3, \infty)\) (d) \((-\infty,-3)\)

The solution set of equation \(3\left[x^{2}+\left(1 / x^{2}\right)\right]+16[x+(1 / x)]+26=0\) is (a) \([-1,\\{(-1) / 3\\},-3]\) (b) \([\overline{1,(1 / 3), 3}]\) (c) \([-1,(1 / 3), 3]\) (d) \([1,\\{(-1) / 3\\}, 3]\)
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