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2x^2+x-1=0 how many real solutions
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Let's solve your equation step-by-step. 2x2+x−1=0 Step 1: Factor left side of equation. (2x−1)(x+1)=0 Step 2: Set factors equal to 0. 2x−1=0 or x+1=0
Answer:x=1/2 or x=−1
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discriminant D=b^2-4ac \[D=1^2-4*2*-1=1+8=9=3^2>0\] roots are real and rational. \[\ because\] it is a qudratic ,so two real roots.
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