Math, asked by chiragmendiratp76b4j, 1 year ago

quadratic polynomial 4x^2+12x+9 has zeros as alfa and beeta. now form a quadratic polynomial whose zeros are alfa-1 and beeta -1

Answers

Answered by bhatiabhupinder17
2

Hope this will be a easy way

Attachments:
Answered by MrInvisible18
13

Answer:

\green {✎﹏﹏﹏﹏﹏﹏﹏﹏﹏﹏}

\blue {Gเѵεɳ \: ƭɦαƭ:-}

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

\red{Tσ \: ƒเɳ∂:-}

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

\green {Sσℓµƭเσɳ:-}

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

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

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

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

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

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

We are given that the zeroes as p,q.

so, p and q must be,

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

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

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

\sf Since, \: p = q

\begin{gathered}\begin{gathered}\\ : \implies \sf \: q - 1 = - \frac{5}{2}\end{gathered}\end{gathered}</h3><p></p><h2>[tex]\purple {Nσω:-}

Sum of the roots is

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

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

The value of product of roots is,

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

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

Similar questions