Question

If \[\alpha \] and \[\beta \] are the zeroes of the polynomial \[3{x^2} + 5x - 2\], then form a quadratic equation polynomial whose zeroes are \[2\alpha \] and \[2\beta \].

Answer 1

Hint: In the given question, \[\alpha \] and \[\beta \] are the zeroes of the polynomial \[3{x^2} + 5x - 2\]. As we know, for a quadratic polynomial \[a{x^2} + bx + c\], sum of its zeroes \[\alpha \] and \[\beta \] is \[ - \dfrac{b}{a}\] and product of zeroes is \[\dfrac{c}{a}\]. Using this we will find the value of \[\left( {\alpha + \beta } \right)\] and \[\alpha \beta \]. Also, the quadratic equation in term of roots is given by: \[{x^2} - \left( {{\text{sum of the roots}}} \right)x + \left( {{\text{product of the roots}}} \right) = 0\]. Using this and substituting the values of \[\left( {\alpha + \beta } \right)\] and \[\alpha \beta \], we will form a quadratic equation polynomial whose zeroes are \[2\alpha \] and \[2\beta \].

Complete step by step solution:
Given, \[\alpha \] and \[\beta \] are the zeroes of the polynomial \[3{x^2} + 5x - 2\] and we have to form a quadratic equation polynomial whose zeroes are \[2\alpha \] and \[2\beta \].
As we know, for a quadratic polynomial \[a{x^2} + bx + c\], sum of its zeroes \[\alpha \] and \[\beta \] is \[ - \dfrac{b}{a}\] and product of zeroes is \[\dfrac{c}{a}\].
So, for the given polynomial \[3{x^2} + 5x - 2\] with zeroes \[\alpha \] and \[\beta \], we get
\[ \Rightarrow \alpha + \beta = - \dfrac{5}{3}\] and \[\alpha \beta = - \dfrac{2}{3}\]
Now we require a quadratic equation whose zeroes are \[2\alpha \] and \[2\beta \].
For this polynomial, we get sum of the roots as:
\[ \Rightarrow 2\alpha + 2\beta = 2\left( {\alpha + \beta } \right)\]
\[ = 2 \times - \dfrac{5}{3}\]
On solving, we get
\[ \Rightarrow 2\alpha + 2\beta = - \dfrac{{10}}{3}\]
Similarly, for this polynomial, we get product of the roots as:
\[ \Rightarrow 2\alpha \times 2\beta = 4\left( {\alpha \beta } \right)\]
\[ = 4 \times \left( { - \dfrac{2}{3}} \right)\]
On solving, we get
\[ \Rightarrow 4\alpha \beta = - \dfrac{8}{3}\]
As we know, a quadratic equation is of form:
\[{x^2} - \left( {{\text{sum of the roots}}} \right)x + \left( {{\text{product of the roots}}} \right) = 0\]
Therefore, a quadratic equation polynomial whose zeroes are \[2\alpha \] and \[2\beta \] is given by
\[ \Rightarrow {x^2} - \left( {2\alpha + 2\beta } \right)x + \left( {4\alpha \beta } \right) = 0\]
Putting the values, we get
\[ \Rightarrow {x^2} + \dfrac{{10}}{3}x - \dfrac{8}{3} = 0\]
Multiplying both the sides by \[3\], we get
\[ \Rightarrow 3{x^2} + 10x - 8 = 0\]

Note:
Zeroes are not affected by multiplying each term of the polynomial by a constant. A quadratic equation function may have one, two, or zero zeroes. Zeroes are also called the x-intercept or roots. Also, the y-coordinate of any points lying on the x-axis is zero. So, to find the zeros of a quadratic function, we set \[f(x) = 0\].