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VirαtKσhli

Given :-

  • The quadratic equation -4x²+mx+n=0 has roots -8 and 1.

To Find :-

  • The value of m and n .

Solution :-

The roots are -8 and 1 . So on substituting x = -8 and 1 , the value of entire polynomial will become 0 . So that ,

[tex]\sf\longrightarrow[/tex] -4x² + mx + n = 0

[tex]\sf\longrightarrow[/tex] -4(-8)² +m(-8) + n = 0

[tex]\sf\longrightarrow[/tex] -256 -8m + n = 0

[tex]\sf\longrightarrow[/tex] 8m - n = -256 ----- (i)

Again ,

[tex]\sf\longrightarrow[/tex] -4(1)² +m(1) + n = 0

[tex]\sf\longrightarrow[/tex] 4 + m + n = 0

[tex]\sf\longrightarrow[/tex] n + m = 4 ----- ( ii)

On adding (i) and (ii) , we have ,

[tex]\sf\longrightarrow[/tex] 9m = 4 - 256

[tex]\sf\longrightarrow[/tex] 9m = -252

[tex]\sf\longrightarrow[/tex] m = -252/9

[tex]\sf\longrightarrow[/tex] m = -28

So the value of n , will be ,

[tex]\sf\longrightarrow[/tex] n = 4-m

[tex]\sf\longrightarrow[/tex] n = 4 - 28 = -24

Hence the required answer is m = -28 and n = -24.

loneguy

[tex]\text{The two roots are,}~~\alpha = -8,~~ \beta = 1\\\\\\\text{Now,}\\\\x^2 - (\alpha + \beta )x+\alpha \beta = 0\\\\\implies x^2 -(-8 +1)x +(-8)(1) =0\\\\\implies x^2+7x-8=0\\\\\text{Multiply both sides by}~ -4 \\\\-4x^2-28x+32=0\\\\\text{By comparing with}~ -4x^2 +mx +n=0, \text{we get,} ~m =-28 ~~ \text{and}~~ n = 32[/tex]

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