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anyone please help me​

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Sagot :

[tex]\\ \sf\longmapsto x^2-4x-5=0[/tex]

  • Use midterm splitation

[tex]\\ \sf\longmapsto x^2-5x+x-5=0[/tex]

[tex]\\ \sf\longmapsto x(x-5)+1(x-5)=0[/tex]

[tex]\\ \sf\longmapsto (x-5)(x+1)=0[/tex]

[tex]\\ \sf\longmapsto x=5\:or\:x=-1[/tex]

Solution is (-1,0) and (5,0)

Graph attached.

Spare way:-

Use quadratic formula .

[tex]\\ \sf\longmapsto x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(-5)}}{2(1)}[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{4\pm \sqrt{16+20}}{2}[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{4\pm 6}{2}[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{10}{2}\:or\:\dfrac{-2}{2}[/tex]

[tex]\\ \sf\longmapsto x=5\:or\:-1[/tex]

View image Аноним

[tex]x {}^{2} - 4x - 5 = 0 \\ x {}^{2} + x - 5x - 5 = 0 \\ x(x + 1) - 5(x + 1) = 0 \\ (x + 1)(x - 5) = 0 \\ x + 1 = 0 \\ x - 5 = 0 \\ x = - 1 \\ x = 5[/tex]

  • The equation has two solutions:

x=5, x=-1

Graph is in the attachment!!

View image Аноним