Question

from quadratic equation whose roots are 3/4and -2/3​

Answer 1

GIVEN :-

• roots of quadratic equation are 3/4 and -2/3.

TO FIND :-

• The quadratic equation by the given roots

SOLUTION :-

As we know that the quadratic equation is given by,

$\\ : \implies \displaystyle \sf \:x^{2} - ( \alpha + \beta )x + ( \alpha \beta ) = 0$

• α = 3/4.
• β = -2/3

Now, by substituting the values we get,

$\\ : \implies \displaystyle \sf \:x^{2} - \bigg( \frac{3}{4} - \frac{2}{3} \bigg)x + \bigg( \frac{3}{4} \times \frac{ - 2}{3} \bigg)= 0$

$: \implies \displaystyle \sf \:x^{2} - \bigg( \frac{9 - 8}{12} \bigg)x + \bigg( \frac{ - 2}{4} \bigg) = 0$

$: \implies \displaystyle \sf \:x^{2} - \frac{1}{12} x + \bigg( \frac{ - 1}{2} \bigg) = 0$

$: \implies \underline{ \boxed{ \displaystyle \sf \bold{ \:x^{2} - \frac{1}{12} x - \frac{1}{2} = 0}}}$

Hence the required quadratic equation is $\displaystyle \sf x^{2} - \frac{1}{12} x - \frac{1}{2} = 0$

Answer 2

Data :

= α = $\frac{3}{4}$

= β = $\frac{-2}{3}$

= $x^2 - ( α + β ) x + ( αβ ) = 0$

Using Values :

= $x^2 - [ ( \frac{3}{4}) + ( \frac{-2}{3}) ] x + [ ( \frac{3}{4}) × ( \frac{-2}{3}) ] = 0$

$x^2 - ( \frac{9 - 8}{12}) x + ( \frac{-2}{4}) = 0$

$x^2 - ( \frac{1}{12}) x + ( \frac{-1}{2}) = 0$

$x^2 - ( \frac{1}{12}) x - ( \frac{1}{2}) = 0$

{ ( + ) × ( - ) = ( - ) }