If the roots of x^2+px+25=0 are equal and both negative, what is the value of p?
Answer and explanation
Correct answer: 10
Let the two equal roots be r and r. Since the equation is monic, the product of the roots is the constant term, so r^2=25. The condition that both roots are negative selects r=-5 rather than r=5. Their sum is therefore r+r=-10. For x^2+px+25=0, the sum of the roots equals -p by Vieta’s relation. Hence -p=-10, which gives p=10. Equivalently, the equation becomes (x+5)^2=x^2+10x+25, confirming the value. Option B would produce roots 5 and 5, which are positive, while 5 and -5 do not form equal roots and do not have product 25 as a repeated pair.
Frequently asked questions
What is the correct answer to this question?
10
Why is this the correct answer?
Let the two equal roots be r and r. Since the equation is monic, the product of the roots is the constant term, so r^2=25. The condition that both roots are negative selects r=-5 rather than r=5. Their sum is therefore r+r=-10. For x^2+px+25=0, the sum of the roots equals -p by Vieta’s relation. Hence -p=-10, which gives p=10. Equivalently, the equation becomes (x+5)^2=x^2+10x+25, confirming the value. Option B would produce roots 5 and 5, which are positive, while 5 and -5 do not form equal roots and do not have product 25 as a repeated pair.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Roots of a Quadratic Equation.
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