Question

Find the discriminant and determine the nature of the roots to the quadratic equation

Answer 1

№17
$x^{2} +6x-7=0 \\ D=6 ^{2} -4*(-7)=36+28=64=8 ^{2} \\ \\ x_{1} = \frac{-6+8}{2} =1 \\ \\ x_{2} = \frac{-6-8}{2} =-7$
Answer: x₁ = 1
               x₂ = - 7

№18
$2 x^{2} -3x+4=3 \\ 2 x^{2} -3x+4-3=0 \\ 2 x^{2} -3x+1=0 \\ \\ D= -3 ^{2} -4*2=9-8=1 \\ \\ x_{1} = \frac{3+1}{2*2} = \frac{4}{4} =1 \\ \\ x_{2} = \frac{3-1}{2*2} = \frac{2}{4} =0.5$
Answer: x₁ = 1
               x₂ = 0,5.

№19
$2 x^{2} -4x-5=0 \\ D=-4 ^{2} -4*2*(-5)=16+40=56 \\ \\ x_{1}= \frac{4+ \sqrt{56} }{2*2} = \frac{4}{4} + \sqrt{ \frac{56}{16} } =1+ \sqrt{3.5} \\ \\ x_{2} = \frac{4- \sqrt{56} }{2*2} = \frac{4}{4} - \sqrt{ \frac{56}{16} } =1- \sqrt{3.5}$
Answer: x₁ = 1 + √3.5
               x₂ = 1 - √3.5