Question

if x=2/3 and x=3 are roots of the quadratic equationax2+7x+b=0, find the values of a and b​

Answer 1

Answer:

3x²-11x+6=0

a=3

b=6

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Answer 2

Answer:

Answer: The value of a and b is 3 and -6 respectively.

Step-by-step explanation:

Since we have given that

$x=\frac{2}{3}\ and\ x=-3x= </p><p>3</p><p>2$

and x=−3 are the roots of the qadratic equation :

$ax^2+7x+b=0ax </p><p>2</p><p> +7x+b=0$

First we put the value of

in the above quadratic equation:

$x=\frac{2}{3}x= </p><p>3</p><p>2$

\begin{gathered}a(\frac{2}{3})^2+7\times \frac{2}{3}+b=0\\\\\frac{4a}{9}+\frac{14}{3}+b=0\\\\\frac{4a}{9}+b=\frac{-14}{3}\\\\4a+9b=-14\times 3=-42\\\\4a+9b=-42-----------(1)\end{gathered}

a(

3

2

$\begin{gathered}a(\frac{2}{3})^2+7\times \frac{2}{3}+b=0\\\\\frac{4a}{9}+\frac{14}{3}+b=0\\\\\frac{4a}{9}+b=\frac{-14}{3}\\\\4a+9b=-14\times 3=-42\\\\4a+9b=-42-----------(1)\end{gathered} </p><p>a( </p><p>3</p><p>2</p><p> </p><p>$

$</p><p>2</p><p> +7× </p><p>3</p><p>2</p><p> </p><p> +b=0</p><p>9</p><p>4a</p><p> </p><p> + </p><p>3</p><p>14</p><p> </p><p> +b=0</p><p>9</p><p>4a</p><p> </p><p> +b= </p><p>3</p><p>−14$

4a+9b=−14×3=−42

4a+9b=−42−−−−−−−−−−−(1)

similarly, if we put the value of x = -3 in the quadratic equation:

$\begin{gathered}-3^2a+7\times -3+b=0\\\\9a-21+b=0\\\\9a+b=21\\\\b=21-9a---(2)\end{gathered} </p><p>−3 </p><p>2$

a+7×−3+b=0

9a−21+b=0

9a+b=21

b=21−9a−−−−−−−−−−−−−(2)

From Eq(1) and (2), we get that

$\begin{gathered}4a+9b=-42\\\\4a+9(21-9a)=-42\\\\4a+189-81a=-42\\\\-77a=-42-189\\\\-77a=-231\\\\a=\frac{231}{77}=3\end{gathered}$

4a+9b=−42

4a+9(21−9a)=−42

4a+189−81a=−42

−77a=−42−189

−77a=−231

$a= </p><p>77</p><p>231</p><p> </p><p> =3$

Now, we put the value of 'a' in Eq(2), we have

$\begin{gathered}b=21-9a\\\\b=21-9\times 3\\\\b=21-27\\\\b=-6\end{gathered}$

b=21−9a

b=21−9×3

b=21−27

b=−6

Hence, the value of a and b is 3 and -6 respectively.