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Question Number 120769 by SOMEDAVONG last updated on 02/Nov/20

$$1).{\mathrm{x}}^2-5=\sqrt{13+\mathrm{x}},\mathrm{Find}\mathrm{x}=?$$$$2).{\mathrm{x}}^3+{3\mathrm{x}}^2-\mathrm{x}-4=0,\mathrm{Find}\mathrm{x}=?$$

Answered by TANMAY PANACEA last updated on 02/Nov/20

$$2)f\left(x\right)=x^3+3x^2-x-4$$$$f\left(0\right)<0$$$$f\left(1\right)<0$$$$f\left(2\right)>0$$$$soonerootliesbetween(1,2)2>x>1$$$$f(-1)<0$$$$f(-2)>0$$$$soonerootliesbetween(-2,-1)$$$$f(-3)<0$$$$soonerootliesbetween(-3,-2)$$

Commented by SOMEDAVONG last updated on 03/Nov/20

$$\mathrm{Thanks}\mathrm{sir}$$

Commented by TANMAY PANACEA last updated on 03/Nov/20

$$mostwelcome$$

Answered by TANMAY PANACEA last updated on 02/Nov/20

$$1)x^2-5=\sqrt{13+x}$$$$x=3isasolution$$$$xcannotbelessthan-13$$$$f\left(x\right)=x^2-5-\sqrt{13+x}$$$$f\left(0\right)<0$$$$f\left(1\right)<0$$$$f\left(2\right)<0$$$$f\left(3\right)=0$$$$f\left(4\right)>0$$$$f\left(5\right)>0$$$$f(-1)<0$$$$f(-2)<0$$$$f(-3)>0$$$$soonerootliesbetween(-3,-2)$$$$$$

Commented by SOMEDAVONG last updated on 03/Nov/20

$$\mathrm{Thanks}\mathrm{sir}$$

Commented by TANMAY PANACEA last updated on 03/Nov/20

$$mostwelcome$$

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