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Chapter 3: Problem 175

If the ratio of the roots of the quadratic equation \(2 x^{2}+16 x+3 k=0\) is \(4: 5\) then \(k=\) (a) \([(2560) /(243)]\) (b) \(\overline{[(243) /(2560)]}\) (c) \([(-2560) /(243)]\) (d) \([(-243) /(2560)]\)

Short Answer

Expert verified
The correct answer is \(k= \frac{(-2560)}{243}\).

Step by step solution

01

Identify the sum and product of the roots

Given the quadratic equation \(2x^2 + 16x + 3k = 0\), let the roots be \(α\) and \(β\). According to the problem, the ratio of the roots is given by \(α:β = 4:5\). Thus, we can express the roots as \(α = 4m\) and \(β = 5m\), where \(m\) is a constant.
02

Apply Vieta's Formulas

Using Vieta's formulas, the sum and product of the roots can be expressed in terms of the coefficients of the quadratic equation as follows: Sum of the roots: \(α + β = \frac{-B}{A}\) Product of the roots: \(α × β = \frac{C}{A}\) Where A, B, and C are the coefficients of the quadratic equation \(2x^2 + 16x + 3k = 0\).
03

Calculate the sum and product of the roots

Using the expressions for the roots \(α = 4m\) and \(β = 5m\), we can calculate the sum and product of the roots: Sum of the roots: \(α + β = 4m + 5m = 9m\) Product of the roots: \(α × β = 4m × 5m = 20m^2\)
04

Set up and solve equations using Vieta's Formulas

Now, we will use Vieta's formulas to set up equations: Sum of the roots: \(\frac{-B}{A} = 9m\) ⟹ \(\frac{-16}{2} = 9m\) Product of the roots: \(\frac{C}{A} = 20m^2\) ⟹ \(\frac{3k}{2} = 20m^2\) Solving the first equation for \(m\): \(-8 = 9m\) \(m = -\frac{8}{9}\) Now, substitute the value of \(m\) in the second equation: \(\frac{3k}{2} = 20\left( -\frac{8}{9} \right)^2\)
05

Solve for k and find the correct answer

Now, we will solve the equation for the value of k: \(3k = 2 × 20\left( -\frac{8}{9} \right)^2\) \(k = \frac{2}{3} × 20\left( -\frac{8}{9} \right)^2\) \(k = \frac{(-2560)}{243}\) Comparing the calculated value of k with the given options, we can see that our answer matches option (c). Therefore, the correct answer is \(k= \boxed{\frac{(-2560)}{243}}\).

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Most popular questions from this chapter

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) \(-6,-1\)

For all \(x \in R\) the number of triplet \((\ell, m, n)\) satisfying equation \(\ell \cos 2 \mathrm{x}+\mathrm{m} \sin ^{2} \mathrm{x}+\mathrm{n}=0\) (a) 2 (b) 4 (c) 6 (d) infinite

If the sum of the two roots of the equation \([1 /(x+a)]+[1 /(x+b)]=(1 / k)\) is zero then their Product is (a) \(\overline{(1 / 2)\left(a^{2}+b^{2}\right)}\) (b) \(\\{(-1) / 2\\}(a+b)^{2}\) (c) \([\\{a+b\\} / 2]^{2}\) (d) None

The number of real values of \(\mathrm{x}\) satisfying the equation \(3\left[x^{2}+\left(1 / x^{2}\right)\right]-16[x+(1 / x)]+26=0\) is (a) 1 (b) 2 (c) 3 (d) 4

If \(\alpha+\beta=5 \alpha^{2}=5 \alpha-3\) and \(\beta^{2}=5 \beta-3\) then the equation whose roots are \((\alpha / \beta)\) and \((\beta / \alpha)\) is (a) \(3 x^{2}-19 x+3=0\) (b) \(x^{2}+5 x-3=0\) (c) \(x^{2}-5 x+3=0\) (d) \(3 x^{2}-25 x+3=0\)
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