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Step-by-step explanation:
1)option:4)Not real
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(1st Question)
If a, and c have opposite signs, then the roots of ax²+bx+c=0 are:
(3) Real and not equal
♦Explanation♦
If a and c have opposite signs, then:
p(x)=ax²+bx–c=0 or –ax²+bx+c=0
Now, For real roots;
Discriminant > 0
i.e., b²–4ac must not be negative
So, For p(x): (b)²–(4)(a)(c)(–1)
"b²+4ac" which must be positive (Thus, Real) , if a and c are real numbers with opposite signs.
Now, For Equal Roots; D=0
Here, For p(x): b²+4ac=0
OR b²=(–4ac)
Where, "b=√[–4ac]", which is not possible(Since, square root of a negative integer is not real), So, this condition is remained unsatisfied.
Which give rise to the thirs option to be correct, i.e.,
3) Real and not Equal
(2nd Question)
If x²–3x+2=0, x²–12x+k=0 have a common root, then k=?
1) 1 or 2
♦Explanation♦
On Finding the roots of the first polynomial:
p₁(x)=x²–3x+2=0
Here, Sum=–3, Product=+2,
Factors=–2,–1
So, p₁(x)=x²–2x–x+2=0
x(x–2)–1(x–2)=0
(x–2)(x–1)=0
∴ x = 2 or x = 1
From the above solution, we can conclude that the roots of the p₁(x) are equal to p₂(x), i.e.,
1) 1 or 2