Question

2x²+5x-7=0
find the roots/ zeros of the above quadratic equation if they exist. if not so find imaginary roots.​

Answer 1

Answer:

Use the quadratic formula

=−±2−4√2

x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}x=2a−b±b2−4ac

Once in standard form, identify a, b and c from the original equation and plug them into the quadratic formula.

22+5−7=0

2x^{2}+5x-7=02x2+5x−7=0

=2

a={\color{#c92786}{2}}a=2

=5

b={\color{#e8710a}{5}}b=5

=−7

c={\color{#129eaf}{-7}}c=−7

=−5±52−4⋅2(−7)√2⋅2

Solution

=1=−72

Step-by-step explanation:

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Answer 2

Answer:

Given equation is x

2

−5x+7=0

Hence, a=1,b=−5,c=7

Therefore discriminant =D=b

2

−4ac

=25−4(7)

=25−28

=−3

Hence, D<0

Therefore, there are no real roots of this equation.

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Step-by-step explanation:

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