Q. Solve the quadratic equation. \(x^2 + 9 = 7x.\)

Q: What is the equation in standard quadratic form?

Rewrite x^2+9=7x by moving terms: x^2-7x+9=0. This is the standard form ax^2+bx+c=0 with a=1,b=-7,c=9.

Q: How do I solve it with the quadratic formula?.

Use x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. Here x=\frac{7\pm\sqrt{49-36}}{2}=\frac{7\pm\sqrt{13}}{2}.

Q: Can it be factored over the integers or rationals?

No. The discriminant is 13, not a perfect square, so it doesn't factor into rational linear factors.

Q: What does the discriminant tell me here?

Discriminant D=b^2-4ac=13>0 so there are two distinct real irrational roots.

Q: What are decimal approximations of the roots?

x\approx\frac{7+\sqrt{13}}{2}\approx 5.3028 and x\approx\frac{7-\sqrt{13}}{2}\approx 1.6972..

Q: What are the vertex and axis of symmetry?

For y=x^2-7x+9, axis is x=\tfrac{7}{2}. Vertex is \left(\tfrac{7}{2},-\tfrac{13}{4}\right). Parabola opens upward..

Q: What are the sum and product of the roots?

By Vieta: sum =7 and product =9, since for x^2-7x+9=0, sum = -b/a=7, product =c/a=9.

Q: How do I check my solutions are correct?

Substitute each root into the original equation x^2+9=7x. If both sides match (within rounding for decimals), the solutions are correct.

Answer

\(x^2-7x+9=0\). By the quadratic formula, \(x=\frac{7\pm\sqrt{49-36}}{2}=\frac{7\pm\sqrt{13}}{2}\).

Detailed Explanation

Solve the quadratic equation

We solve the equation \(x^{2} + 9 = 7x\) with a step-by-step detailed explanation.

Rewrite the equation in standard quadratic form (move all terms to one side):
\[x^{2} + 9 = 7x\]

Subtract \(7x\) from both sides to obtain:

\[x^{2} – 7x + 9 = 0\]
Identify the coefficients of the quadratic \(ax^{2} + bx + c = 0\):
\[a = 1,\quad b = -7,\quad c = 9\]
Compute the discriminant \(\Delta = b^{2} – 4ac\). The discriminant determines the nature of the roots:
\[\Delta = (-7)^{2} – 4\cdot 1 \cdot 9 = 49 – 36 = 13\]

Since \(\Delta = 13 > 0\), the equation has two distinct real roots.
Apply the quadratic formula \(x = \dfrac{-b \pm \sqrt{\Delta}}{2a}\):
\[x = \frac{-(-7) \pm \sqrt{13}}{2\cdot 1} = \frac{7 \pm \sqrt{13}}{2}\]
Write the two solutions explicitly:
\[x = \frac{7 + \sqrt{13}}{2}\quad\text{and}\quad x = \frac{7 – \sqrt{13}}{2}\]

Optional decimal approximations:

\[\sqrt{13}\approx 3.605551275\]

\[x \approx \frac{7 + 3.605551275}{2} \approx 5.302775638\]

\[x \approx \frac{7 – 3.605551275}{2} \approx 1.697224362\]

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Algebra FAQs

What is the equation in standard quadratic form?

Rewrite \(x^2+9=7x\) by moving terms: \(x^2-7x+9=0\). This is the standard form \(ax^2+bx+c=0\) with \(a=1,b=-7,c=9\).

How do I solve it with the quadratic formula?.

Use \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Here \(x=\frac{7\pm\sqrt{49-36}}{2}=\frac{7\pm\sqrt{13}}{2}\).

Can it be factored over the integers or rationals?

No. The discriminant is \(13\), not a perfect square, so it doesn't factor into rational linear factors.

What does the discriminant tell me here?

Discriminant \(D=b^2-4ac=13>0\) so there are two distinct real irrational roots.

What are decimal approximations of the roots?

\(x\approx\frac{7+\sqrt{13}}{2}\approx 5.3028\) and \(x\approx\frac{7-\sqrt{13}}{2}\approx 1.6972\)..

How do I solve by completing the square?

What are the vertex and axis of symmetry?

For \(y=x^2-7x+9\), axis is \(x=\tfrac{7}{2}\). Vertex is \(\left(\tfrac{7}{2},-\tfrac{13}{4}\right)\). Parabola opens upward..

What are the sum and product of the roots?

By Vieta: sum \(=7\) and product \(=9\), since for \(x^2-7x+9=0\), sum \(= -b/a=7\), product \(=c/a=9\).

How do I check my solutions are correct?

Substitute each root into the original equation \(x^2+9=7x\). If both sides match (within rounding for decimals), the solutions are correct.

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