Use the discriminant to describe the roots of each equation. Then select the best description. 1. x^2 - 5x + 7 = 0 2. x^2 - 5x - 4 = 0 3. 16x^2 + 8x + 1 = 0 4. x^2 + 9x + 14 = 0
here are the options a.double root b.real and rational root c.real and irrational root d.imaginary root
do you know how to use the discriminant..?
no not really
The discriminant is \[b^2 - 4ac\] if the discriminant is positive, there are 2 real roots if it is negative, then there are 2 imaginary roots if it is 0, then there is one real root
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ok... so for a quadratic \[ax^2 + bx + c\] the discriminant is \[\Delta = b^2 - 4ac\] in your 1st question a = 1, b = -5 and c = 7 can you substitute the values and calculate the value of the discriminant..?
the value will determine the type of roots your quadratic equation has \[\Delta > 0\] is 2 two unequal real roots if the descriminant is a perfect square such as 1, 4, 9, 16, ... the the roots are rational... otherwise there are irrational. \[\Delta = 0\] the two roots are equal.... e.f x^2 - 4x + 4 = (x-2)^2 and has 2 equal roots of x = 2 \[\Delta < 0\] the roots are unreal... or complex... hope it helps
ok so can you show me how you would solve #1
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@ShadowLegendX
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ok... in 1... a = 1 b = -5 and c = 7 using \[\Delta = b^2 - 4ac\] so \[\Delta = (-5)^2 - 4 \times 1 \times 7\] \[\Delta = 25 - 28\] so \[\Delta= -3\] this is less than zero... so it has complex or imaginary roots... hope it helps
what about #2?
well a = 1, b = -5 and c = -4 so you can substitute them and see what you get...
okay so i think its a.double root
well its 2 unequal roots that are irrational... this is because the answer is 39...
what would this one be? x^2 - 6x + 12 = 0
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