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Question Number 20053 by Tinkutara last updated on 21/Aug/17

$$\mathrm{If}{(4a+c)}^2⩽4b^2\mathrm{then}\mathrm{one}\mathrm{root}\mathrm{of}$$$${ax}^2+bx+c=0\mathrm{lies}\mathrm{in}$$$$\left(1\right)(-2,2)$$$$\left(2\right)(-1,1)$$$$\left(3\right)(-\mathrm{\infty },-2)$$$$\left(4\right)(2,\mathrm{\infty })$$

Answered by ajfour last updated on 21/Aug/17

$${(4a+c)}^2-4b^2⩽0$$$$\Rightarrow (4a+2b+c)(4a-2b+c)⩽0$$$$\Rightarrow f\left(2\right)f(-2)⩽0$$$$\Rightarrow f\left(x\right)changessignoriszero$$$$in[-2,2],henceoneroot$$$$liesin[-2,2].$$

Commented by Tinkutara last updated on 21/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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