Question

If

2/3

and −

1/2

are solutions of quadratic equation 6x2 + ax − b = 0; find the values of a and b.​

Answer 1

Solution,

Given,

x= $-\frac{1}{2}$    {for positive(+) in quadratic equation}

x= $\frac{2}{3}$        {for negative(-) in quadratic equation}

equation: $6x^2+ax-b=0$

(a, b)=?

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now,

comparing $6x^2+ax-b=0$ with $ax^2+bx+c=0$, we get;

a= 6, b=a, c=-b

Then,

Applying quadratic equation formula, we get;

$x=\frac{-b \frac{+}{} (\sqrt{b^2-4ac})} {2a}$

  = $\frac{-a\frac{+}{}(\sqrt{a^2-4*6*(-b)})}{2*6}$

 =$\frac{-a\frac{+}{}(\sqrt{a^2+24b})}{12}$

now ,

taking(+), we get;

x = $\frac{-a+(\sqrt{a^2+24b})}{12}$

or, $-\frac{1}{2}$ = $\frac{-a+(\sqrt{a^2+24b})}{12}$           [when x=$-\frac{1}{2}$]

or, -6 = $-a+(\sqrt{a^2+24b})}$

or, a-6= $(\sqrt{a^2+24b})$

squaring on both side, we get;

or, $(a-6)^2 = (\sqrt{a^2+24b})^2$

or, $a^2-2*a*6+6^2= a^2+24b$

or, 36-12a= 24b

or12(3-a)=12*2b

or, 3-a=2b

:. a=3-2b----------------------(I)

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Taking(-), we get;

x= $\frac{-a-(\sqrt{a^2+24b})}{12}$

or, $\frac{2}{3}=\frac{-a+(\sqrt{a^2+24b})}{12}$                         [ when x= $\frac{2}{3}$]

or, 8= $-[a+(\sqrt{a^2+24b})]$

or,-8= a+$(\sqrt{a^2+24b})$

or,-8-a=$(\sqrt{a^2+24b})$

squaring on both side, we get;

or, $(-8-a)^2=(\sqrt{a^2+24b})^2$

or, 64+2*8*a+$a^2= a^2+24b$

or, 64+16a=24b

or, 8(8+2a)=8*3b

or, 8+2a=3b

or, 8+2(3-2b)=3b

or, 8+6-4b=3b

or, 14=4b+3b

or, 14=7b

:. b= 2

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Again,

Inserting value of b in equation (I), we get;

a=3-2*2

 =3-4

:.a= -1

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Thus, (a, b) is (-1, 2)

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