Question

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Answer 1

Step-by-step explanation:

1)option:4)Not real

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Answer 2

(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