Question

The roots of the equation 9x^2-bx +18 =o will be equal ,if the value of b is

Answer 1

$Compare \: given \: Quadratic \: equation$

$9x^{2} - bx + 18 = 0\:with \: Ax^{2}+Bx +C=0,$

$we \:get$

$A = 9 , B = -b , C = 18$

$Discreminant (D) = 0 \: [ Given \:equal \:roots]$

$\implies B^{2} - 4AC = 0$

$\implies (-b)^{2} = 4AC$

$\implies b^{2} = 4\times 9 \times 18$

$\implies b^{2} = 2\times 2 \times 9 \times 9 \times 2$

$\implies b= \pm \sqrt{2\times 2 \times 9 \times 9 \times 2}$

$\implies b= \pm 2\times 9 \sqrt{2}$

$\implies b= \pm 18 \sqrt{2}$

Therefore.,

$\red{ Value \:of \:b } \green{= \pm 18 \sqrt{2}}$

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Answer 2

Step-by-step explanation:

ComparegivenQuadraticequation

9x^{2} - bx + 18 = 0\:with \: Ax^{2}+Bx +C=0,9x

2

−bx+18=0withAx

2

+Bx+C=0,

we \:getweget

A = 9 , B = -b , C = 18A=9,B=−b,C=18

Discreminant (D) = 0 \: [ Given \:equal \:roots]Discreminant(D)=0[Givenequalroots]

\implies B^{2} - 4AC = 0⟹B

2

−4AC=0

\implies (-b)^{2} = 4AC⟹(−b)

2

=4AC

\implies b^{2} = 4\times 9 \times 18⟹b

2

=4×9×18

\implies b^{2} = 2\times 2 \times 9 \times 9 \times 2⟹b

2

=2×2×9×9×2

\implies b= \pm \sqrt{2\times 2 \times 9 \times 9 \times 2}⟹b=±

2×2×9×9×2

\implies b= \pm 2\times 9 \sqrt{2}⟹b=±2×9

2

\implies b= \pm 18 \sqrt{2}⟹b=±18

2

Therefore.,