Question

x^2-ax -21=0 and x² – 3ax +35=0, a>0, have a common root, then a is

Answer 1

Step-by-step explanation:

Let be the common roots of:

² – ℎ – 21 = 0

and

² – 3ℎ + 35 = 0

Then by definition of a root, we have the system:

{ ² – ℎ – 21 = 0

{ ² – 3ℎ + 35 = 0

► Subtracting second equation from the first yields:

(² – ℎ – 21) – (² – 3ℎ + 35) = 0 – 0

² – ℎ – 21 – ² + 3ℎ – 35 = 0

² – ² – ℎ + 3ℎ – 21 – 35 = 0

0 + 2ℎ – 56 = 0

2ℎ = 56

= 28/ℎ

► Inserting this in the first equation:

² – ℎ – 21 = 0

(28/ℎ)² – ℎ(28/ℎ) – 21 = 0 ← Since =28/ℎ

784/ℎ² – 28 – 21 = 0

784/ℎ² – 49 = 0

784/ℎ² = 49

16/ℎ² = 1 ← Divide both sides by 49

16 = ℎ² ← Multiply both sides by ℎ²

ℎ = ±4

► Now since we are told that ℎ>0, the solution ℎ=-4 is discarded leading to the final conclusion that:

ℎ = 4 ◄ANSWER

Answer 2

Answer: a = 4.

Given: $x^2-ax -21=0$ and $x^2-3ax +35=0$.

To Find: The value of a.

Step-by-step explanation:

Step 1: Quadratic equations are polynomial equations of degree two in a single variable of the form $f(x) = ax^2 + bx + c = 0$ where a, b, c, and an are all divisible by R and an is equal to zero. It is the generic form of a quadratic equation, where "a" is the leading coefficient and "c" is the absolute term of f(x). The quadratic equation's roots (, ) correspond to values of x that fulfil the equation in four variables.

Step 2: Two roots are required for the quadratic equation. There are two possible types of roots: actual and made-up.Let the common root is α. Then it will satisfied both the given equation. Thus, $x^2-ax -21=0$ and $x^2-3ax +35=0$

Put the common root in both the equation we get,

$\alpha^2-a\alpha -21=0 \ ...eq(1)$

$\alpha ^2-3a\alpha +35=0\ ...eq(2)$

From equation(1) and (2) we get,

$-2a\alpha+56=0\\\\\alpha =\frac{28}{a}$

Step 3: Now put the value of α in equation(1). We get

$(\frac{28}{a} )^2-a(\frac{28}{a}) -21=0\\\\(\frac{28}{a} )^2-28 -21=0\\\\(\frac{28}{a} )^2-49=0\\\\(\frac{28}{a} )=\pm 7\\\\a=\pm 4$

Here a > 0, thus a = 4.

Hence, a = 4 is the correct answer.

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