Find the value of the product of roots of this quadratic equation It is given that one of the roots of the quadratic equation : x2+(p+3)x−p2=0

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Find the value of the product of roots of this quadratic equation  It is given that one of the roots of the quadratic equation :  x2+(p+3)x−p2=0

coolbananas03ok · Accepted Answer

Step 1 Without applying the factorization of a &quot;difference of two squares&quot; or Viete&#039;s relations, we can still use the information stated in the problem. If we call the two roots of the quadratic equation r and -r , then we have r2+(p+3)×r−p2=0 and [−r]2+(p+3)×[−r]−p2 =r2−(p+3)×r−p2=0 This means that r2=−(p+3)×r+p2 =(p+3)×r+p2⇒2×(p+3)×r=0 So either r=0 or p=−3 But if r=0=−r, then 02+(p+3)×0−p2=0 would reqiure p=0, which would then make the quadratic equation x3+3×x=0 But that polynomial factors as x×(x+3)=0, so we couldn&#039;t have both roots equal to zero. Instead, it must be that p=−3, making the equation x2+0×x−(−3)2=x2−9=0, for which the roots are given by r2=9⇒r=+3, −3; the product of the roots is thus -9 Step 2 Another way to arrive at this conclusion is that y=x2+(p+3)×x−p2 is the equation of an &quot;upward-opening&quot; parabola, for which we want the x-intercepts to be x=−r and x=r. Its axis of symmetry is located midway between these intercepts, so we have h=0 in the &quot;vertex form&quot; of the parabola&#039;s equation, y=(x−0)2−p2 (The vertex is definitely &quot;below&quot; the x-axis at (0, −p2), so we know these x-intercepts exist.) The equation of the parabola is therefore y=x2−p2, making p+3=0 and the rest of the argument above follows.

lirwerwammete9t · Accepted Answer

Step 1We know that this equation has at most two roots in the set of reals.Let&#039;s denote them with a and -a. Then your equation isx2−a2=0.Therefore p must be equal to -3.Hence the equation isx2−9=0and the roots are 3 and -3. So the desired product is -9.

Find the value of the product of roots of this

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