Please solve 6th one : 6 Solve for x : x 2 - 1 + x - 1 2 + x 2 - 3 x + 2 = 0

Dear student x2-1+x-12+x2-3x+2=0x2-1=-x-12-x2-3x+2x2-1=-x2+2x-1-x2-3x+2Now testing each absolute for its positive and negative ranges x2-1≥0x2≥1Now, fx2≥a⇒fx≥a or fx≤-ax≥1 or x≤-1Now, x2-1 -ax -1So, final solution is :-1<x<1Evaluate the expression in the following ranges:x<-1, -1≤x<1, x≥1For , x<-1Replacing x2-1 with x2-1x2-1+x-12+x2-3x+2=0x2-1+x2+1-2x+x2-3x+2=02x2-2x+x2-3x+2=0x2-3x+2=-2x2+2xx2-3x+22=-2x2+2x2x2-3x+2=4x4+4x2-8x34x4-8x3+3x2+3x-2=04x4-4x3-x2+2x-4x3+4x2+x-2=0x4x3-4x2-x+2-14x3-4x2-x+2=0x-14x3-4x2-x+2=0x-1=0 4x3-4x2-x+2=0 we cannot find the roots To find the roots we need Newton Raphson method which is not a part of your syllabus x=1 Regards

Please solve 6th one....:

6. Solve for $x : |x^{2} - 1| + {(x - 1)}^{2} + \sqrt{x^{2} - 3 x + 2} = 0$

Dear student

|x^{2} - 1| + {(x - 1)}^{2} + \sqrt{x^{2} - 3 x + 2} = 0 |x^{2} - 1| = - {(x - 1)}^{2} - \sqrt{x^{2} - 3 x + 2} |x^{2} - 1| = - x^{2} + 2 x - 1 - \sqrt{x^{2} - 3 x + 2} Now testing each absolute for its positive and negative ranges . x^{2} - 1 \geq 0 x^{2} \geq 1 Now, {(f (x))}^{2} \geq a \Rightarrow f (x) \geq \sqrt{a} or f (x) \leq - \sqrt{a} x \geq 1 or (x \leq - 1) Now, x^{2} - 1 < 0 x^{2} < 1 Now, {(f (x))}^{2} < a \Rightarrow f (x) < \sqrt{a} and f (x) > - \sqrt{a} x < 1 and (x > - 1) So, final solution is : - 1 < x < 1 Evaluate the expression in the following ranges : x < - 1, - 1 \leq x < 1, x \geq 1 For, x < - 1 Replacing |x^{2} - 1| with (x^{2} - 1) x^{2} - 1 + {(x - 1)}^{2} + \sqrt{x^{2} - 3 x + 2} = 0 x^{2} - 1 + x^{2} + 1 - 2 x + \sqrt{x^{2} - 3 x + 2} = 0 2 x^{2} - 2 x + \sqrt{x^{2} - 3 x + 2} = 0 \sqrt{x^{2} - 3 x + 2} = - 2 x^{2} + 2 x {(\sqrt{x^{2} - 3 x + 2})}^{2} = {(- 2 x^{2} + 2 x)}^{2} x^{2} - 3 x + 2 = 4 x^{4} + 4 x^{2} - 8 x^{3} 4 x^{4} - 8 x^{3} + 3 x^{2} + 3 x - 2 = 0 4 x^{4} - 4 x^{3} - x^{2} + 2 x - 4 x^{3} + 4 x^{2} + x - 2 = 0 x (4 x^{3} - 4 x^{2} - x + 2) - 1 (4 x^{3} - 4 x^{2} - x + 2) = 0 (x - 1) (4 x^{3} - 4 x^{2} - x + 2) = 0 x - 1 = 0 [(4 x^{3} - 4 x^{2} - x + 2) = 0 we cannot find the roots . To find the roots we need Newton Raphson method which is not a part of your syllabus .]

x = 1

Regards

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Please solve 6th one....: 6. Solve for x : x 2 - 1 + x - 1 2 + x 2 - 3 x + 2 = 0

Please solve 6th one....:

6. Solve for $x : |x^{2} - 1| + {(x - 1)}^{2} + \sqrt{x^{2} - 3 x + 2} = 0$