Question 1

Anyone please help me

Question 2

$x {}^(2) - 4x - 5 = 0 \\ x {}^(2) + x - 5x - 5 = 0 \\ x(x + 1) - 5(x + 1) = 0 \\ (x + 1)(x - 5) = 0 \\ x + 1 = 0 \\ x - 5 = 0 \\ x = - 1 \\ x = 5$

Question 3

$\\ \sf\longmapsto x^2-4x-5=0$

$\\ \sf\longmapsto x^2-5x+x-5=0$

$\\ \sf\longmapsto x(x-5)+1(x-5)=0$

$\\ \sf\longmapsto (x-5)(x+1)=0$

$\\ \sf\longmapsto x=5\:or\:x=-1$

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

Graph attached.

Spare way:-

Use quadratic formula .

$\\ \sf\longmapsto x=(-b\pm √(b^2-4ac))/(2a)$

$\\ \sf\longmapsto x=(-(-4)\pm√((-4)^2-4(1)(-5)))/(2(1))$

$\\ \sf\longmapsto x=(4\pm √(16+20))/(2)$

$\\ \sf\longmapsto x=(4\pm 6)/(2)$

$\\ \sf\longmapsto x=(10)/(2)\:or\:(-2)/(2)$

$\\ \sf\longmapsto x=5\:or\:-1$

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