Question

Quadratic polynomial 4x^2+12x+9 has zeroes as p and q. Now form a Quadratic polynomial whose zeroes are p-1 and q-1

Answer 1

Answer:

hi

Step-by-step explanation:

Sol: Given α and β are zeroes of the polynomial f(x) = x2- 4x + 3 α+ β = 4 αβ = 3 1) (3α + 3β) = 3x 4 = 12 3α x 3β = 9 x 3 = 27. If 3α, 3β are zeros of the quadratic polynomial then the equation is x2 -(3α + 3β)x + 9αβ = 0 then x2 - 12x + 27 = 0. 2) (1/2α + 1/2β) = (α + β) / 2αβ = 4 / 6 = 2 / 3. 1/4αβ = 1 /12 If 1 / 2α, 1 / 2β are zeros of the quadratic polynomial then the equation is x2 -(1 / 2α + 1 / 2β)x + 1 / 4αβ = 0 then x2 -(2 / 3)x + 1 / 12 = 0 12x2 - 8x + 1 = 0.

Answer 2

Answer:

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A quadratic polynomial 4x² + 12x + 9 has zeroes p and q

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The quadratic polynomial whose zeroes are p - 1 and q - 1.

$\pink {Tɦεɳ:-}$

The qudratic equation when its roots α , β are given is given by,

$\begin{gathered}\\ : \implies \sf \: x^2-(\alpha + \beta)x+\alpha\beta \\ \\\end{gathered}$

$\green {Sσℓµƭเσɳ:-}$

First of all let us find the zeroes of 4x² +12x + 9.

$\begin{gathered}\\ : \implies \sf \: 4 {x}^{2} + 12x + 9 = 0 \\ \\\end{gathered}$

$\begin{gathered}\\ :\implies \sf \: 4 {x}^{2} + 6x + 6x + 9 = 0 \\ \\\end{gathered}$

$\begin{gathered}\\ : \implies \sf \: 2x(2x + 3) + 3(2x + 3) = 0 \\ \\\end{gathered}$

$\begin{gathered}\\ : \implies \sf \: (2x + 3)(2x + 3) = 0 \\ \\\end{gathered}$

$\begin{gathered}\\ : \implies \sf \: x = - </h3><h3>\frac{3}{2}\end{gathered}$

We are given that the zeroes as p,q.

so, p and q must be,

$\begin{gathered}\\ : \implies \sf \: p = - \frac{3}{2} \: and \: q = - \frac{3}{2}\end{gathered}$

We are given that other quadratic equation has p - 1 and q - 1 as roots.

Then the values of roots of that quadratic equations are,

$\begin{gathered}\\ : \implies \sf \: p - 1 = \frac{ - 3}{2} - 1 \\ \\\end{gathered}$

$\begin{gathered}\\ : \implies \sf \: p - 1 = \frac{ - 5}{2} \: \\ \\\end{gathered}$

The value of q - 1 is same as p - 1.

$\sf Since, \: p = q$

$\begin{gathered}\\ : \implies \sf \: q - 1 = - \frac{5}{2}\end{gathered}$

$\purple {Nσω:-}$

Sum of the roots is

$\begin{gathered}\\ : \implies \sf \:( p - 1)+( q - 1) = \frac{ - 5}{2} + \frac{ - 5}{2} \\ \\\end{gathered}$

$\begin{gathered}\\ : \implies \sf \: (p - 1) + (q - 1) = \frac{ - 10}{2} = -5\end{gathered}$

The value of product of roots is,

$\begin{gathered}\\ : \implies \sf \: (p - 1)(q - 1)= \frac{ - 5}{2} \times \frac{ - 5}{2} \\ \\\end{gathered}$

$\begin{gathered}\\ : \implies \sf \: (p - 1)(q - 1) = \frac{25}{4}\end{gathered}$

The required quadratic equation is,

$\begin{gathered}\\ : \implies \sf \: {x}^{2} -(-5)x + \dfrac{25}{4}=0\end{gathered}$

Multiplying the whole equation with 4 then we get,

$\begin{gathered}\\ :\implies\underline{\boxed{\sf\:{4x^2+20x+25 =0}}}\\ \\\end{gathered}$

$\orange {Hεɳcє:-}$

The Required quadratic equation is 4x² + 20x + 25

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